MathsCBSE Class 10 · Chapter 8

Introduction to Trigonometry Class 10: Concepts, Formulas and Methods

Q: Are trigonometric ratios of complementary angles in the 2026-27 syllabus?

No. Trigonometric ratios of complementary angles were removed in the NCERT rationalisation and are not in the 2026-27 syllabus.

Q: Should I memorise the table of values or derive it?

Do both. Memorise the table so you are fast in MCQs, but know how it comes from the isosceles right triangle and the half equilateral triangle, so you can rebuild any value you forget.

Q: How do I start proving an identity when I feel stuck?

Pick the more complicated side and convert every ratio into sin and cos. Take the LCM, look for sin squared plus cos squared, and factorise. If one side still will not simplify, work on the other side separately until both match.

Q: Why is tan 90 degrees not defined?

Tan is sin divided by cos, and cos 90 degrees is 0. Division by zero has no meaning, so tan 90 degrees is not defined. The same reason makes sec 90 degrees, cot 0 degrees and cosec 0 degrees undefined.

Q: How is this chapter connected to heights and distances?

The next chapter, Some Applications of Trigonometry, uses exactly these ratios and standard values to find heights of towers and distances between objects. If you can label a right triangle and recall the values at 30, 45 and 60 degrees confidently, that chapter becomes much easier.

Updated 13 October 2026 · Fact-checked

Introduction to Trigonometry in Class 10 studies the six trigonometric ratios of an acute angle in a right-angled triangle, their values at 0, 30, 45, 60 and 90 degrees, and identities built on sin squared A plus cos squared A equals 1. The board asks you to find ratios, evaluate expressions, find angles and prove simple identities.

What Introduction to Trigonometry is about

Trigonometry means measuring triangles. In this chapter you take a right-angled triangle, fix one acute angle, and compare its sides in pairs. These comparisons are the six trigonometric ratios: $sin$ , $cos$ , $tan$ , $cosec$ , $sec$ and $cot$ . You learn why each ratio depends only on the angle, the exact values at $0^{\circ}$ , $3 0^{\circ}$ , $4 5^{\circ}$ , $6 0^{\circ}$ and $9 0^{\circ}$ , and the identity $sin^{2} A + cos^{2} A = 1$ with its two partner identities.

The chapter rests on two things you already know: Pythagoras theorem, and similar triangles from the Triangles chapter. It leads straight into Some Applications of Trigonometry, where you use these same ratios to find heights and distances. If your ratios and standard values are solid, that chapter becomes easy.

Board questions come in a few clear shapes. One-mark MCQs and assertion-reason items test standard values, which ratios are undefined at $0^{\circ}$ or $9 0^{\circ}$ , and quick identity use. Two- and three-mark questions ask you to find all ratios from one given ratio, evaluate an expression, find an unknown angle, or prove a simple identity. Longer questions combine an identity with a given condition, and case studies put a right triangle into a real setting such as a ramp or a ladder.

Weightage: This chapter is part of Unit V, Trigonometry, which carries 12 of the 80 theory marks together with Some Applications of Trigonometry.Source: CBSE Mathematics syllabus 2026-27, Unit V Trigonometry

In real board papers: Standard, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 6.7 marks per paper; Basic, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 6.6 marks per paper. See the questions

Key concepts

Trigonometric ratios of an acute angle

NCERT section 8.2 Trigonometric Ratios

Take a right-angled triangle $A B C$ with $∠ B = 9 0^{\circ}$ and look at the acute angle $A$ . Relative to $A$ , side $B C$ is the side opposite, $A B$ is the side adjacent, and $A C$ is the hypotenuse. The six ratios compare these sides in pairs: $sin A = \frac{B C}{A C}$ , $cos A = \frac{A B}{A C}$ , $tan A = \frac{B C}{A B}$ , and their reciprocals $cosec A = \frac{A C}{B C}$ , $sec A = \frac{A C}{A B}$ , $cot A = \frac{A B}{B C}$ .

Opposite and adjacent always depend on which angle you stand at. For $∠ C$ the opposite side becomes $A B$ . So mark the angle first, then label the sides.

Use Pythagoras, $A C^{2} = A B^{2} + B C^{2}$ , whenever a side is missing. The hypotenuse is the longest side, so $sin A$ and $cos A$ of an acute angle always lie between 0 and 1. Remember that $sin A$ is one symbol; it is not $sin \times A$ .

Worked example

In $△ P QR$ , right-angled at $Q$ , $P Q = 12$ cm and $P R + QR = 18$ cm. Find $sin P$ , $cos P$ , $tan P$ and $tan R$ .

Try it first, then show the solutionSolution

$∠ Q = 9 0^{\circ}$ , so $P R$ is the hypotenuse.
By Pythagoras, $P R^{2} = P Q^{2} + Q R^{2}$ , so $P R^{2} - Q R^{2} = 144$ .
Factorise: $(P R - QR) (P R + QR) = 144$ .
Substitute $P R + QR = 18$ : $P R - QR = \frac{144}{18} = 8$ .
Add the two equations: $2 P R = 26$ , so $P R = 13$ cm and $QR = 18 - 13 = 5$ cm.
For $∠ P$ : opposite side $QR = 5$ , adjacent side $P Q = 12$ , hypotenuse $P R = 13$ .
$sin P = \frac{5}{13}$ , $cos P = \frac{12}{13}$ , $tan P = \frac{5}{12}$ .
For $∠ R$ : opposite side is $P Q = 12$ , adjacent side is $QR = 5$ , so $tan R = \frac{12}{5}$ .
Check: $sin^{2} P + cos^{2} P = \frac{25 + 144}{169} = 1$ .

Answer: $sin P = \frac{5}{13}$ , $cos P = \frac{12}{13}$ , $tan P = \frac{5}{12}$ , $tan R = \frac{12}{5}$ (sides $P R = 13$ cm, $QR = 5$ cm).

Why the ratios are well defined

NCERT section 8.2 Trigonometric Ratios

Does $sin A$ change if you draw a bigger triangle? No, and here is why. Take two right triangles $A B C$ and $A^{'} B^{'} C^{'}$ , both right-angled at $B$ and $B^{'}$ , with $∠ A = ∠ A^{'}$ . Two pairs of angles are equal, so by AA similarity $△ A B C \sim △ A^{'} B^{'} C^{'}$ . Corresponding sides of similar triangles are proportional, so $\frac{B C}{A C} = \frac{B ^{'} C ^{'}}{A ^{'} C ^{'}}$ . That means $sin A$ is the same number in both triangles. The same argument works for every ratio.

So each trigonometric ratio depends only on the angle, not on the size of the triangle. This is what makes the ratios well defined.

It also gives you a working tool. If $sin A = \frac{3}{5}$ , you cannot say $B C = 3$ and $A C = 5$ for certain; you write $B C = 3 k$ and $A C = 5 k$ for some positive $k$ . The $k$ always cancels when you form another ratio.

Worked example

Assertion (A): In $△ A B C$ , right-angled at $B$ , if every side is doubled to form a new right triangle with the same angles, the value of $tan A$ does not change.
Reason (R): The trigonometric ratios of an angle depend only on the angle and not on the lengths of the sides of the triangle.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

New triangle has sides $2 A B$ , $2 B C$ , $2 A C$ , so it is similar to $△ A B C$ (SSS similarity).
In the new triangle, $tan A = \frac{2 B C}{2 A B} = \frac{B C}{A B}$ , the same as before. So A is true.
R is the general fact, proved using similar triangles, that ratios depend only on the angle. So R is true.
A is a direct case of R, so R correctly explains A.

Answer: Option (a): Both A and R are true and R is the correct explanation of A.

Finding all ratios from one given ratio

NCERT section 8.2 Trigonometric Ratios

When one ratio is given, you can find all the others. Write the given ratio as sides with a constant $k$ , find the third side by Pythagoras, then read off whatever you need.

For example, if $tan θ = \frac{5}{12}$ , take opposite $= 5 k$ and adjacent $= 12 k$ . Then the hypotenuse is $25 k^{2} + 144 k^{2} = 13 k$ , so $sin θ = \frac{5}{13}$ and $cos θ = \frac{12}{13}$ . The $k$ cancels every time.

When you must evaluate an expression like $\frac{a s i n θ + b c o s θ}{c s i n θ + d c o s θ}$ , there is a shortcut: divide top and bottom by $cos θ$ to get everything in $tan θ$ . Use it as a check on your answer. If $θ$ is acute, all ratios are positive, so you never take a negative square root.

Worked example

If $sec θ = \frac{13}{12}$ , where $θ$ is acute, find the value of $\frac{2 s i n θ - 3 c o s θ}{4 s i n θ - 9 c o s θ}$ .

Try it first, then show the solutionSolution

$sec θ = \frac{13}{12}$ , so $cos θ = \frac{12}{13}$ : hypotenuse $= 13 k$ , adjacent $= 12 k$ .
Opposite side $= 169 k^{2} - 144 k^{2} = 5 k$ (Pythagoras).
So $sin θ = \frac{5}{13}$ .
Numerator: $2 \cdot \frac{5}{13} - 3 \cdot \frac{12}{13} = \frac{10 - 36}{13} = - \frac{26}{13}$ .
Denominator: $4 \cdot \frac{5}{13} - 9 \cdot \frac{12}{13} = \frac{20 - 108}{13} = - \frac{88}{13}$ .
Value $= \frac{- 26}{- 88} = \frac{13}{44}$ .
Check with $tan θ = \frac{5}{12}$ : $\frac{2 t a n θ - 3}{4 t a n θ - 9} = \frac{( 10 - 36 ) /12}{( 20 - 108 ) /12} = \frac{26}{88} = \frac{13}{44}$ .

Answer: $\frac{13}{44}$

Values of the ratios at 30°, 45° and 60°

NCERT section 8.3 Trigonometric Ratios of Some Specific Angles

You can build these values from two simple triangles, so you never need to guess.

** $4 5^{\circ}$ :** a right triangle with $∠ A = 4 5^{\circ}$ is isosceles. Take both legs $a$ ; the hypotenuse is $a 2$ . So $sin 4 5^{\circ} = cos 4 5^{\circ} = \frac{1}{2}$ and $tan 4 5^{\circ} = 1$ .

** $3 0^{\circ}$ and $6 0^{\circ}$ :** take an equilateral triangle of side $2 a$ and drop a perpendicular from the top vertex. It cuts the base into two parts of length $a$ and has length $4 a^{2} - a^{2} = 3 a$ . Each half is a right triangle with angles $3 0^{\circ}$ , $6 0^{\circ}$ , $9 0^{\circ}$ and sides $a$ , $3 a$ , $2 a$ . So $sin 3 0^{\circ} = \frac{1}{2}$ , $cos 3 0^{\circ} = \frac{3}{2}$ , $tan 3 0^{\circ} = \frac{1}{3}$ , and $sin 6 0^{\circ} = \frac{3}{2}$ , $cos 6 0^{\circ} = \frac{1}{2}$ , $tan 6 0^{\circ} = 3$ .

When you square a value, square the whole fraction: $cos^{2} 3 0^{\circ} = \frac{3}{4}$ , not $\frac{3}{2}$ .

Worked example

Evaluate: $\frac{3 t a n ^{2} 6 0 ^{\circ} + 4 c o s ^{2} 4 5 ^{\circ} - 2 s e c ^{2} 3 0 ^{\circ}}{cosec ^{2} 3 0 ^{\circ} - c o t ^{2} 4 5 ^{\circ}}$

Try it first, then show the solutionSolution

$tan 6 0^{\circ} = 3$ , so $3 tan^{2} 6 0^{\circ} = 3 \times 3 = 9$ .
$cos 4 5^{\circ} = \frac{1}{2}$ , so $4 cos^{2} 4 5^{\circ} = 4 \times \frac{1}{2} = 2$ .
$sec 3 0^{\circ} = \frac{2}{3}$ , so $2 sec^{2} 3 0^{\circ} = 2 \times \frac{4}{3} = \frac{8}{3}$ .
Numerator $= 9 + 2 - \frac{8}{3} = \frac{33 - 8}{3} = \frac{25}{3}$ .
$cosec 3 0^{\circ} = 2$ and $cot 4 5^{\circ} = 1$ , so denominator $= 4 - 1 = 3$ .
Value $= \frac{25}{3} \div 3 = \frac{25}{9}$ .

Answer: $\frac{25}{9}$

Using standard values to find unknown angles

NCERT section 8.3 Trigonometric Ratios of Some Specific Angles

The table also works backwards. As an angle grows from $0^{\circ}$ to $9 0^{\circ}$ , $sin$ increases steadily from 0 to 1 and $cos$ decreases from 1 to 0. So each value belongs to exactly one angle in that range. If $cos θ = \frac{1}{2}$ and $θ$ is acute, then $θ = 6 0^{\circ}$ and nothing else.

Board questions use this in two ways. First, an equation such as $2 sin 3 x = 3$ gives $sin 3 x = \frac{3}{2}$ , so $3 x = 6 0^{\circ}$ and $x = 2 0^{\circ}$ . Second, two conditions on $A + B$ and $A - B$ give two linear equations in $A$ and $B$ , which you solve by adding and subtracting.

This also shows why $sin (A + B)$ is not $sin A + sin B$ . With $A = B = 3 0^{\circ}$ : $sin 6 0^{\circ} = \frac{3}{2}$ , but $sin 3 0^{\circ} + sin 3 0^{\circ} = 1$ .

Worked example

If $sin (A + B) = 1$ and $cos (A - B) = \frac{3}{2}$ , where $0^{\circ} < A + B \leq 9 0^{\circ}$ and $A > B$ , find $A$ and $B$ . Hence find the value of $tan A \cdot cot B$ .

Try it first, then show the solutionSolution

$sin (A + B) = 1 = sin 9 0^{\circ}$ , so $A + B = 9 0^{\circ}$ ... (1)
$cos (A - B) = \frac{3}{2} = cos 3 0^{\circ}$ , so $A - B = 3 0^{\circ}$ ... (2) (positive because $A > B$ )
Add (1) and (2): $2 A = 12 0^{\circ}$ , so $A = 6 0^{\circ}$ .
From (1): $B = 9 0^{\circ} - 6 0^{\circ} = 3 0^{\circ}$ .
Check: $sin 9 0^{\circ} = 1$ and $cos 3 0^{\circ} = \frac{3}{2}$ .
$tan A \cdot cot B = tan 6 0^{\circ} \cdot cot 3 0^{\circ} = 3 \times 3 = 3$ .

Answer: $A = 6 0^{\circ}$ , $B = 3 0^{\circ}$ , and $tan A \cdot cot B = 3$ .

Ratios at 0° and 90° (and which are not defined)

NCERT section 8.3 Trigonometric Ratios of Some Specific Angles

Picture $△ A B C$ right-angled at $B$ , with hypotenuse $A C$ fixed in length. Make $∠ A$ smaller and smaller. Side $B C$ (opposite $A$ ) shrinks towards 0 and $A B$ grows towards $A C$ . So $sin A = \frac{B C}{A C}$ gets close to 0 and $cos A = \frac{A B}{A C}$ gets close to 1. This motivates $sin 0^{\circ} = 0$ , $cos 0^{\circ} = 1$ , $tan 0^{\circ} = 0$ , $sec 0^{\circ} = 1$ .

Now make $∠ A$ grow towards $9 0^{\circ}$ . Then $A B$ shrinks to 0 and $B C$ approaches $A C$ . So $sin 9 0^{\circ} = 1$ , $cos 9 0^{\circ} = 0$ , $cot 9 0^{\circ} = 0$ , $cosec 9 0^{\circ} = 1$ .

Division by zero is not allowed. So $cosec 0^{\circ}$ and $cot 0^{\circ}$ are not defined (they need $sin 0^{\circ} = 0$ in the denominator), and $tan 9 0^{\circ}$ and $sec 9 0^{\circ}$ are not defined (they need $cos 9 0^{\circ} = 0$ in the denominator). Write "not defined", never 0 or infinity.

Worked example

If $θ = 0^{\circ}$ , which of the following is defined?
(a) $cosec θ$
(b) $cot θ$
(c) $sec θ + tan θ$
(d) $\frac{1}{1 - c o s θ}$

Try it first, then show the solutionSolution

(a) $cosec 0^{\circ} = \frac{1}{s i n 0 ^{\circ}} = \frac{1}{0}$ : not defined.
(b) $cot 0^{\circ} = \frac{c o s 0 ^{\circ}}{s i n 0 ^{\circ}} = \frac{1}{0}$ : not defined.
(c) $sec 0^{\circ} + tan 0^{\circ} = 1 + 0 = 1$ : defined.
(d) $\frac{1}{1 - c o s 0 ^{\circ}} = \frac{1}{1 - 1} = \frac{1}{0}$ : not defined.

Answer: Option (c): $sec θ + tan θ$ , whose value at $0^{\circ}$ is 1.

Relationships between the ratios

NCERT section 8.2 Trigonometric Ratios

The six ratios are not independent. Three are simply reciprocals of the other three: $cosec A = \frac{1}{s i n A}$ , $sec A = \frac{1}{c o s A}$ , $cot A = \frac{1}{t a n A}$ . You can see this straight from the sides: $sin A = \frac{B C}{A C}$ and $cosec A = \frac{A C}{B C}$ .

There is also a quotient link. Divide top and bottom of $tan A = \frac{B C}{A B}$ by $A C$ : $tan A = \frac{B C / A C}{A B / A C} = \frac{s i n A}{c o s A}$ . Similarly $cot A = \frac{c o s A}{s i n A}$ .

These links are your main tool for simplifying. When an expression mixes several ratios, rewrite everything in $sin$ and $cos$ , take the LCM, and simplify. This works because every other ratio is built from these two.

Worked example

Prove that $\frac{1 + t a n A}{1 + c o t A} = tan A$ , where $sin A \neq = 0$ , $cos A \neq = 0$ and $sin A + cos A \neq = 0$ . Check your result for $A = 6 0^{\circ}$ .

Try it first, then show the solutionSolution

LHS $= \frac{1 + \frac{s i n A}{c o s A}}{1 + \frac{c o s A}{s i n A}}$ (quotient relations).
Numerator $= \frac{c o s A + s i n A}{c o s A}$ ; denominator $= \frac{s i n A + c o s A}{s i n A}$ .
LHS $= \frac{c o s A + s i n A}{c o s A} \times \frac{s i n A}{s i n A + c o s A}$ .
Cancel $(sin A + cos A)$ : LHS $= \frac{s i n A}{c o s A} = tan A =$ RHS.
Check at $A = 6 0^{\circ}$ : $\frac{1 + 3}{1 + \frac{1}{3}} = \frac{( 1 + 3 ) 3}{3 + 1} = 3 = tan 6 0^{\circ}$ .

Answer: LHS simplifies to $tan A$ , so the identity is proved; at $A = 6 0^{\circ}$ both sides equal $3$ .

The identity sin²A + cos²A = 1 and its two partners

NCERT section 8.4 Trigonometric Identities

In $△ A B C$ right-angled at $B$ , Pythagoras gives $A B^{2} + B C^{2} = A C^{2}$ . Divide every term by $A C^{2}$ : $(\frac{A B}{A C})^{2} + (\frac{B C}{A C})^{2} = 1$ , which is $cos^{2} A + sin^{2} A = 1$ . Using the values at $0^{\circ}$ and $9 0^{\circ}$ , it holds for $0^{\circ} \leq A \leq 9 0^{\circ}$ .

Divide the same Pythagoras equation by $A B^{2}$ instead: $1 + tan^{2} A = sec^{2} A$ , true for $0^{\circ} \leq A < 9 0^{\circ}$ . Divide by $B C^{2}$ : $cot^{2} A + 1 = cosec^{2} A$ , true for $0^{\circ} < A \leq 9 0^{\circ}$ .

Learn the rearranged forms too: $cos^{2} A = 1 - sin^{2} A$ , $sec^{2} A - tan^{2} A = 1$ , and the factorised form $(sec A - tan A) (sec A + tan A) = 1$ . The factorised form is the key to questions that give you $sec θ + tan θ$ or $cosec θ + cot θ$ .

Worked example

If $sec θ + tan θ = p$ (with $θ$ acute), find $sec θ - tan θ$ in terms of $p$ , and hence prove that $sin θ = \frac{p ^{2} - 1}{p ^{2} + 1}$ .

Try it first, then show the solutionSolution

Given $sec θ + tan θ = p$ ... (1)
Since $sec^{2} θ - tan^{2} θ = 1$ , we have $(sec θ - tan θ) (sec θ + tan θ) = 1$ .
So $sec θ - tan θ = \frac{1}{p}$ ... (2)
Add (1) and (2): $2 sec θ = p + \frac{1}{p} = \frac{p ^{2} + 1}{p}$ , so $sec θ = \frac{p ^{2} + 1}{2 p}$ .
Subtract (2) from (1): $2 tan θ = p - \frac{1}{p}$ , so $tan θ = \frac{p ^{2} - 1}{2 p}$ .
$\frac{t a n θ}{s e c θ} = \frac{s i n θ / c o s θ}{1/ c o s θ} = sin θ$ .
So $sin θ = \frac{p ^{2} - 1}{2 p} \times \frac{2 p}{p ^{2} + 1} = \frac{p ^{2} - 1}{p ^{2} + 1}$ . Hence proved.
Check with $θ = 6 0^{\circ}$ : $p = 2 + 3$ , $p^{2} = 7 + 43$ , and $\frac{6 + 4 3}{8 + 4 3}$ simplifies to $\frac{3}{2} = sin 6 0^{\circ}$ .

Answer: $sec θ - tan θ = \frac{1}{p}$ and $sin θ = \frac{p ^{2} - 1}{p ^{2} + 1}$ .

Proving simple trigonometric identities

NCERT section 8.4 Trigonometric Identities

An identity is true for every allowed angle, so you cannot treat it like an equation and move terms across the equals sign. Instead, start with one side (usually the more complicated one) and transform it step by step until it becomes the other side.

A reliable method:

Convert $tan$ , $cot$ , $sec$ , $cosec$ into $sin$ and $cos$ .
Combine fractions using the LCM.
Replace $1 - sin^{2} A$ by $cos^{2} A$ (or similar) using the identities.
Factorise, especially $a^{2} - b^{2} = (a - b) (a + b)$ , and cancel common factors.

If you get stuck, work on the other side separately and show both reach the same expression. State any condition needed for cancelling, for example $cos A \neq = sin A$ .

Worked example

Prove that $\frac{c o s A}{1 - t a n A} + \frac{s i n A}{1 - c o t A} = sin A + cos A$ , where $A$ is acute and $A \neq = 4 5^{\circ}$ .

Try it first, then show the solutionSolution

LHS $= \frac{c o s A}{1 - \frac{s i n A}{c o s A}} + \frac{s i n A}{1 - \frac{c o s A}{s i n A}}$
First term $= \frac{c o s A}{\frac{c o s A - s i n A}{c o s A}} = \frac{c o s ^{2} A}{c o s A - s i n A}$ .
Second term $= \frac{s i n ^{2} A}{s i n A - c o s A} = - \frac{s i n ^{2} A}{c o s A - s i n A}$ .
LHS $= \frac{c o s ^{2} A - s i n ^{2} A}{c o s A - s i n A}$ .
$= \frac{( c o s A - s i n A ) ( c o s A + s i n A )}{c o s A - s i n A}$ (difference of squares).
$= cos A + sin A =$ RHS, since $cos A - sin A \neq = 0$ for $A \neq = 4 5^{\circ}$ .

Answer: LHS reduces to $sin A + cos A$ , so the identity is proved.

Maths in real life: Introduction to Trigonometry

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Finding all ratios from one given ratio

Imagine your family is putting up Diwali lights on the outside wall of your house. You use a ladder 6.5 m long, leaning against the wall on level ground. The ladder's instructions say it is safest when the angle $θ$ it makes with the ground satisfies $tan θ = \frac{12}{5}$ .

At this safe angle, how high up the wall does the top of the ladder reach, and how far from the wall is its foot? A window ledge where the lights must be fixed is 5.8 m above the ground. Can the ladder reach it at this angle? Also find $sin θ$ and $cos θ$ .

Try it first, then show the solutionSolution

The wall, the ground and the ladder form a right triangle, right-angled at the foot of the wall.
$tan θ = \frac{12}{5}$ , so take height on the wall (opposite) $= 12 k$ and distance of foot from the wall (adjacent) $= 5 k$ , where $k > 0$ .
Hypotenuse $= (12 k)^{2} + (5 k)^{2} = 169 k^{2} = 13 k$ .
The hypotenuse is the ladder, so $13 k = 6.5$ , giving $k = 0.5$ .
Height reached $= 12 k = 12 \times 0.5 = 6$ m.
Distance of the foot from the wall $= 5 k = 5 \times 0.5 = 2.5$ m.
$sin θ = \frac{12 k}{13 k} = \frac{12}{13}$ and $cos θ = \frac{5 k}{13 k} = \frac{5}{13}$ .
Check: $6^{2} + 2. 5^{2} = 36 + 6.25 = 42.25 = 6. 5^{2}$ . The sides are correct.
Since $6$ m $> 5.8$ m, the ladder reaches the window ledge.

Answer: The top reaches 6 m up the wall and the foot is 2.5 m from the wall. Yes, the ladder reaches the 5.8 m ledge. $sin θ = \frac{12}{13}$ , $cos θ = \frac{5}{13}$ .

Memory hook: One ratio given? Write the sides as $12 k$ and $5 k$ , find $13 k$ by Pythagoras, then let the real ladder length fix $k$ .

Values of the ratios at 30°, 45° and 60°

Suppose a park near your home is installing a children's slide. The straight sliding surface is 4 m long and makes an angle of $3 0^{\circ}$ with the level ground. At the top it meets a straight ladder that makes an angle of $6 0^{\circ}$ with the ground on the other side. The slide and the ladder lie in the same vertical plane, and the top of the slide is joined to the ground by a vertical support pole.

Find the height of the top of the slide above the ground, the length of the ladder, and the total distance along the ground from the foot of the ladder to the bottom end of the slide. Leave answers in surd form, then give approximate values using $3 \approx 1.732$ .

Try it first, then show the solutionSolution

Let $T$ be the top and $P$ the point on the ground directly below $T$ . Then $T P$ is the height, and $∠ T P S = ∠ T P L = 9 0^{\circ}$ , where $S$ is the bottom of the slide and $L$ is the foot of the ladder.
In right $△ T P S$ : $sin 3 0^{\circ} = \frac{T P}{T S}$ , so $T P = 4 \times \frac{1}{2} = 2$ m.
$cos 3 0^{\circ} = \frac{P S}{T S}$ , so $P S = 4 \times \frac{3}{2} = 23$ m.
In right $△ T P L$ : $sin 6 0^{\circ} = \frac{T P}{T L}$ , so $T L = \frac{2}{3 /2} = \frac{4}{3} = \frac{4 3}{3}$ m.
$tan 6 0^{\circ} = \frac{T P}{P L}$ , so $P L = \frac{2}{3} = \frac{2 3}{3}$ m.
The ladder and slide are on opposite sides of $P$ , so $L S = P L + P S = \frac{2 3}{3} + 23 = \frac{2 3 + 6 3}{3} = \frac{8 3}{3}$ m.
Approximate values: $T L \approx \frac{4 \times 1.732}{3} \approx 2.31$ m and $L S \approx \frac{8 \times 1.732}{3} \approx 4.62$ m.
Check for the ladder: $P L^{2} + T P^{2} = \frac{4}{3} + 4 = \frac{16}{3} = T L^{2}$ . Correct.

Answer: Height $= 2$ m; ladder length $= \frac{4 3}{3}$ m $\approx 2.31$ m; ground distance from the foot of the ladder to the bottom of the slide $= \frac{8 3}{3}$ m $\approx 4.62$ m.

Memory hook: The park slide: $3 0^{\circ}$ means half the length goes up ( $sin 3 0^{\circ} = \frac{1}{2}$ ).

Using standard values to find unknown angles

Imagine you are on a school camping trip, helping to put up a tent. A vertical tent pole is 1.5 m tall. A straight guy rope runs from the top of the pole to a peg in the level ground. At first the rope is 3 m long.

Find the angle the rope makes with the ground. The teacher then shortens the rope to $1.5 2$ m and moves the peg so that the rope is tight again. Find the new angle with the ground, and find how much closer to the pole the peg has moved. Use $3 \approx 1.732$ for the final value.

Try it first, then show the solutionSolution

The pole, the ground and the rope form a right triangle, right-angled at the foot of the pole. Let $θ$ be the angle between the rope and the ground.
First position: $sin θ = \frac{pole}{rope} = \frac{1.5}{3} = \frac{1}{2}$ .
Since $θ$ is acute and $sin 3 0^{\circ} = \frac{1}{2}$ , $θ = 3 0^{\circ}$ .
Distance of the first peg from the pole $= 3 cos 3 0^{\circ} = 3 \times \frac{3}{2} = \frac{3 3}{2}$ m.
Second position: $sin θ^{'} = \frac{1.5}{1.5 2} = \frac{1}{2}$ , so $θ^{'} = 4 5^{\circ}$ .
Distance of the new peg from the pole $= 1.5 2 \times cos 4 5^{\circ} = 1.5 2 \times \frac{1}{2} = 1.5$ m.
Shift of the peg $= \frac{3 3}{2} - 1.5 = 1.5 (3 - 1)$ m.
Approximately $1.5 \times 0.732 = 1.098 \approx 1.1$ m.
Check: $1. 5^{2} + 1. 5^{2} = 4.5 = (1.5 2)^{2}$ . Correct.

Answer: The first angle is $3 0^{\circ}$ and the new angle is $4 5^{\circ}$ . The peg moves $1.5 (3 - 1)$ m $\approx 1.1$ m closer to the pole.

Memory hook: Tent-rope trick: a rope twice the pole's height means $sin θ = \frac{1}{2}$ , so the angle is $3 0^{\circ}$ .

The identity sin²A + cos²A = 1 and its two partners

Suppose a furniture shop uses a straight loading plank to roll a cupboard up into the back of a delivery van. The plank's manual says that when the plank is placed correctly, the angle $A$ it makes with the ground has $sec A = 1.25$ . The foot of the plank is 2 m from the van, measured along level ground.

Using an identity, find $tan A$ . Then find the height of the van's loading floor above the ground and the length of the plank. Finally, find $sin A$ and $cos A$ , and check that $sin^{2} A + cos^{2} A = 1$ .

Try it first, then show the solutionSolution

The ground, the vertical back of the van and the plank form a right triangle, right-angled below the loading edge. The horizontal run, 2 m, is adjacent to $A$ .
Use $1 + tan^{2} A = sec^{2} A$ : $tan^{2} A = 1.2 5^{2} - 1 = 1.5625 - 1 = 0.5625$ .
$A$ is acute, so $tan A$ is positive: $tan A = 0.75 = \frac{3}{4}$ .
Height $= run \times tan A = 2 \times \frac{3}{4} = 1.5$ m.
Plank length $= run \times sec A = 2 \times 1.25 = 2.5$ m, because $sec A = \frac{hypotenuse}{adjacent}$ .
$cos A = \frac{1}{s e c A} = \frac{1}{1.25} = \frac{4}{5}$ , and $sin A = \frac{1.5}{2.5} = \frac{3}{5}$ .
Check: $sin^{2} A + cos^{2} A = \frac{9}{25} + \frac{16}{25} = 1$ . Correct.
Pythagoras also agrees: $1. 5^{2} + 2^{2} = 2.25 + 4 = 6.25 = 2. 5^{2}$ .

Answer: $tan A = \frac{3}{4}$ ; height of the loading floor $= 1.5$ m; plank length $= 2.5$ m; $sin A = \frac{3}{5}$ , $cos A = \frac{4}{5}$ , and $sin^{2} A + cos^{2} A = 1$ is satisfied.

Memory hook: Loading plank: know $sec$ , get $tan$ from $sec^{2} A - tan^{2} A = 1$ , which is just Pythagoras divided by the run squared.

Topics in this chapter

Introduction to Trigonometry formula sheet

Sine of an acute angle: $sin A = \frac{side opposite A}{hypotenuse}$; Label sides from the angle you are working with; the opposite side changes if you switch angles.
Cosine of an acute angle: $cos A = \frac{side adjacent to A}{hypotenuse}$; Adjacent is the side next to the angle that is not the hypotenuse.
Tangent of an acute angle: $tan A = \frac{side opposite A}{side adjacent to A}$; The only one of the three basic ratios that does not use the hypotenuse; it can be greater than 1.
Reciprocal relations: $cosec A = \frac{1}{s i n A}, sec A = \frac{1}{c o s A}, cot A = \frac{1}{t a n A}$; Each pair has exactly one 'co': sin with cosec, cos with sec, tan with cot.
Quotient relations: $tan A = \frac{s i n A}{c o s A}, cot A = \frac{c o s A}{s i n A}$; Use these to convert everything to sin and cos when simplifying or proving identities.
Pythagoras theorem (right angle at B): $A C^{2} = A B^{2} + B C^{2}$; Use to find the third side when a ratio or two sides are given.
Sine values: $sin 0^{\circ} = 0, sin 3 0^{\circ} = \frac{1}{2}, sin 4 5^{\circ} = \frac{1}{2}, sin 6 0^{\circ} = \frac{3}{2}, sin 9 0^{\circ} = 1$; Increases from 0 to 1 as the angle goes from 0° to 90°.
Cosine values: $cos 0^{\circ} = 1, cos 3 0^{\circ} = \frac{3}{2}, cos 4 5^{\circ} = \frac{1}{2}, cos 6 0^{\circ} = \frac{1}{2}, cos 9 0^{\circ} = 0$; The sine row written in reverse order.
Tangent values: $tan 0^{\circ} = 0, tan 3 0^{\circ} = \frac{1}{3}, tan 4 5^{\circ} = 1, tan 6 0^{\circ} = 3, tan 9 0^{\circ} not defined$; Get each by dividing sin by cos for that angle.
Cosecant values: $cosec 0^{\circ} not defined, cosec 3 0^{\circ} = 2, cosec 4 5^{\circ} = 2, cosec 6 0^{\circ} = \frac{2}{3}, cosec 9 0^{\circ} = 1$; Reciprocals of the sine row.
Secant values: $sec 0^{\circ} = 1, sec 3 0^{\circ} = \frac{2}{3}, sec 4 5^{\circ} = 2, sec 6 0^{\circ} = 2, sec 9 0^{\circ} not defined$; Reciprocals of the cosine row.
Cotangent values: $cot 0^{\circ} not defined, cot 3 0^{\circ} = 3, cot 4 5^{\circ} = 1, cot 6 0^{\circ} = \frac{1}{3}, cot 9 0^{\circ} = 0$; Reciprocals of the tangent row.
Side ratios of special triangles: $3 0^{\circ} - 6 0^{\circ} - 9 0^{\circ} : 1 : 3 : 2, 4 5^{\circ} - 4 5^{\circ} - 9 0^{\circ} : 1 : 1 : 2$; Sides opposite the angles in the order listed; rebuild any standard value from these.
Fundamental identity: $sin^{2} A + cos^{2} A = 1, 0^{\circ} \leq A \leq 9 0^{\circ}$; Also use as cos²A = 1 − sin²A and sin²A = 1 − cos²A.
Identity with sec and tan: $1 + tan^{2} A = sec^{2} A, 0^{\circ} \leq A < 9 0^{\circ}$; Divide the fundamental identity by cos²A; also sec²A − tan²A = 1.
Identity with cosec and cot: $1 + cot^{2} A = cosec^{2} A, 0^{\circ} < A \leq 9 0^{\circ}$; Divide the fundamental identity by sin²A; also cosec²A − cot²A = 1.
Factorised forms: $(sec A - tan A) (sec A + tan A) = 1, (cosec A - cot A) (cosec A + cot A) = 1$; If sec A + tan A = p, then sec A − tan A = 1/p. Same idea for cosec and cot.
Bounds for an acute angle: $0 < sin A < 1, 0 < cos A < 1, sec A > 1, cosec A > 1 (0^{\circ} < A < 9 0^{\circ})$; Quick check: a value like sin A = 4/3 is impossible.
Caution: ratios do not split over sums: $sin (A + B) \neq = sin A + sin B$; Always find the angle A + B first, then take its ratio.

How the board asks Introduction to Trigonometry

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Find other ratios from one given ratio	Given something like 3 tan θ = 4 or sec θ = 13/12, find the remaining ratios or evaluate an expression in sin θ and cos θ. Appears as MCQs and 2-mark questions.	Write the given ratio as sides $ak$ and $bk$ , find the third side by Pythagoras, and list the ratios. For an expression in $sin$ and $cos$ of the same degree, divide top and bottom by $cos θ$ and use $tan θ$ as a check.
Right triangle with a condition on its sides	A right triangle is described with one side known and the sum or difference of the hypotenuse and another side given; find the sides and then the ratios of an angle.	Write Pythagoras as a difference of squares, $h^{2} - s^{2} = (h - s) (h + s)$ , substitute the given sum or difference, and solve the two linear equations. Then label sides relative to the asked angle.
Evaluate an expression using standard values	An expression mixing ratios of 0°, 30°, 45°, 60° and 90°, often with squares, to be simplified to a number. Common in Sections A and B.	Replace each ratio by its exact value, square whole fractions carefully, simplify numerator and denominator separately, then divide. Rationalise only if the question asks.
Find an unknown angle or variable	Equations like 2 sin 3x = √3, or tan 2x equal to a combination of standard values, or two conditions on A + B and A − B.	Simplify to a single ratio equal to a standard value, read off the acute angle, then solve for the variable. For two conditions, form two linear equations in $A$ and $B$ and add or subtract. Always substitute back to check.
Defined or not, and reasoning items	MCQs or assertion-reason items on which ratios are undefined at 0° or 90°, whether a value like sin A = 5/4 is possible, or whether a claimed result is true.	Recall that only $cot, cosec$ fail at $0^{\circ}$ and only $tan, sec$ fail at $9 0^{\circ}$ . Use bounds: $sin$ and $cos$ of an acute angle lie between 0 and 1. For a claim, test it with a standard angle.
Prove a simple identity	A 2- or 3-mark proof: show that a given expression in trigonometric ratios equals another.	Start from the more complicated side. Convert to $sin$ and $cos$ , take the LCM, use $sin^{2} A + cos^{2} A = 1$ or its partners, factorise and cancel. Never move terms across the equals sign.
Identity with a given condition	Given a relation such as sec θ + tan θ = p or sin θ + cos θ = √2, prove a result or find another expression. Can carry 3 to 5 marks.	Look for a factorised identity or square the given relation. Example: if $sin θ + cos θ = 2$ , squaring gives $1 + 2 sin θ cos θ = 2$ , so $sin θ cos θ = \frac{1}{2}$ and $tan θ + cot θ = \frac{1}{s i n θ c o s θ} = 2$ .
Case study on a right triangle	Section E style: a ramp, ladder, roof or slide is described with some lengths; sub-parts ask for a ratio, a missing side, and an expression or identity check.	Draw the right triangle from the description, mark the right angle and the angle in question, label opposite, adjacent and hypotenuse, then apply the definitions and Pythagoras. Write units in every length.

Tricks to remember Introduction to Trigonometry

Short memory aids. Say each one aloud once, then use it on a question the same day.

SOH CAH TOA
Sin = Opposite/Hypotenuse, Cos = Adjacent/Hypotenuse, Tan = Opposite/Adjacent. Read it as three short words: SOH, CAH, TOA.
Use it when: Writing the three basic ratios correctly from any labelled triangle.
Square-root-over-2 pattern for sine
For $0^{\circ}, 3 0^{\circ}, 4 5^{\circ}, 6 0^{\circ}, 9 0^{\circ}$ write $\frac{0}{2}, \frac{1}{2}, \frac{2}{2}, \frac{3}{2}, \frac{4}{2}$ . These simplify to $0, \frac{1}{2}, \frac{1}{2}, \frac{3}{2}, 1$ . Cosine is the same row backwards.
Use it when: Rebuilding the whole sine and cosine table in seconds.
Tan row by division
Divide each sine value by the cosine value of the same angle: $0, \frac{1}{3}, 1, 3$ , not defined. Then flip the tan row to get cot.
Use it when: Getting tan and cot values without memorising them separately.
Undefined pairs
At $0^{\circ}$ , $sin = 0$ , so the ratios with $sin$ underneath fail: $cosec$ and $cot$ . At $9 0^{\circ}$ , $cos = 0$ , so $sec$ and $tan$ fail.
Use it when: Answering defined-or-not MCQs and assertion-reason items quickly.
Sum-of-squares check
After finding $sin θ$ and $cos θ$ , check that their squares add to 1. For example $(\frac{5}{13})^{2} + (\frac{12}{13})^{2} = 1$ .
Use it when: Catching a wrong side or arithmetic slip before moving on.
Identity proof routine: S-C-L-F
Sin-cos (convert), Combine (LCM), Look for identities, Factorise and cancel.
Use it when: A fixed sequence that gets you started on any simple identity.

Introduction to Trigonometry: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Value from a condition on the ratios · 1 mark · MCQBoard level

If $sin θ = cos θ$ for an acute angle $θ$ , then the value of $2 tan^{2} θ + sin^{2} θ - 1$ is
(a) $\frac{1}{2}$
(b) $\frac{3}{2}$
(c) $2$
(d) $\frac{5}{2}$

Try it first, then show the solutionSolution

Divide $sin θ = cos θ$ by $cos θ$ , which is not zero for an acute angle: $tan θ = 1$ , so $θ = 4 5^{\circ}$ .
$tan^{2} 4 5^{\circ} = 1$ and $sin^{2} 4 5^{\circ} = \frac{1}{2}$ .
$2 (1) + \frac{1}{2} - 1 = \frac{3}{2}$

Answer: (b) $\frac{3}{2}$

Example 2: Does sin split over a sum? · 1 mark · Assertion-reasonBoard level

Assertion (A): For $A = 6 0^{\circ}$ and $B = 3 0^{\circ}$ , $sin (A + B) = sin A + sin B$ .
Reason (R): As $θ$ increases from $0^{\circ}$ to $9 0^{\circ}$ , the value of $sin θ$ increases.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

LHS: $sin (6 0^{\circ} + 3 0^{\circ}) = sin 9 0^{\circ} = 1$ .
RHS: $sin 6 0^{\circ} + sin 3 0^{\circ} = \frac{3}{2} + \frac{1}{2} = \frac{3 + 1}{2}$ , which is more than 1 because $3 > 1$ . So A is false.
R is true: $sin θ$ takes the values $0, \frac{1}{2}, \frac{1}{2}, \frac{3}{2}, 1$ at $0^{\circ}, 3 0^{\circ}, 4 5^{\circ}, 6 0^{\circ}, 9 0^{\circ}$ , so it increases.

Answer: (d) A is false but R is true.

Example 3: Right triangle from a ratio and a sum of sides · 3 marks · Short answerHarder board level

In $△ A B C$ , right-angled at $B$ , $tan C = \frac{3}{4}$ and $A B + A C = 24$ cm. Find the three sides, then find $sin A$ , $cos C$ and $tan A$ . Hence evaluate $sin A cos C + cos A sin C$ .

Try it first, then show the solutionSolution

At $C$ , the opposite side is $A B$ and the adjacent side is $B C$ . So $tan C = \frac{A B}{B C} = \frac{3}{4}$ : take $A B = 3 k$ and $B C = 4 k$ , $k > 0$ . (½ mark)
By Pythagoras theorem, $A C = 9 k^{2} + 16 k^{2} = 5 k$ . (½ mark)
$A B + A C = 3 k + 5 k = 8 k = 24$ , so $k = 3$ . Hence $A B = 9$ cm, $B C = 12$ cm, $A C = 15$ cm. (½ mark)
At $A$ , the opposite side is $B C$ and the adjacent side is $A B$ : $sin A = \frac{12}{15} = \frac{4}{5}$ , $tan A = \frac{12}{9} = \frac{4}{3}$ . At $C$ : $cos C = \frac{B C}{A C} = \frac{4}{5}$ . (1 mark)
$cos A = \frac{9}{15} = \frac{3}{5}$ and $sin C = \frac{A B}{A C} = \frac{3}{5}$ , so $sin A cos C + cos A sin C = \frac{16}{25} + \frac{9}{25} = 1$ . (½ mark)
Check: $9^{2} + 1 2^{2} = 81 + 144 = 225 = 1 5^{2}$ , and $9 + 15 = 24$ .

Answer: $A B = 9$ cm, $B C = 12$ cm, $A C = 15$ cm; $sin A = \frac{4}{5}$ , $cos C = \frac{4}{5}$ , $tan A = \frac{4}{3}$ ; the expression equals 1.

Example 4: Product of three brackets equals 1 · 3 marks · Short answerBoard level

Prove that $(cosec A - sin A) (sec A - cos A) (tan A + cot A) = 1$ , where $A$ is acute. Verify the result for $A = 4 5^{\circ}$ .

Try it first, then show the solutionSolution

$cosec A - sin A = \frac{1}{s i n A} - sin A = \frac{1 - s i n ^{2} A}{s i n A} = \frac{c o s ^{2} A}{s i n A}$ , using $sin^{2} A + cos^{2} A = 1$ . (1 mark)
$sec A - cos A = \frac{1}{c o s A} - cos A = \frac{1 - c o s ^{2} A}{c o s A} = \frac{s i n ^{2} A}{c o s A}$ . (½ mark)
$tan A + cot A = \frac{s i n A}{c o s A} + \frac{c o s A}{s i n A} = \frac{s i n ^{2} A + c o s ^{2} A}{s i n A c o s A} = \frac{1}{s i n A c o s A}$ . (1 mark)
LHS $= \frac{c o s ^{2} A}{s i n A} \times \frac{s i n ^{2} A}{c o s A} \times \frac{1}{s i n A c o s A} = \frac{s i n A c o s A}{s i n A c o s A} = 1 =$ RHS. (½ mark)
Check at $A = 4 5^{\circ}$ : $2 - \frac{1}{2} = \frac{1}{2}$ for each of the first two brackets, and $tan 4 5^{\circ} + cot 4 5^{\circ} = 2$ . So the product is $\frac{1}{2} \times \frac{1}{2} \times 2 = 1$ .

Answer: LHS reduces to 1, so the identity is proved; at $A = 4 5^{\circ}$ the product is also 1.

Example 5: Replacing 1 by an identity · 3 marks · Short answerHarder board level

Prove that $\frac{t a n θ + s e c θ - 1}{t a n θ - s e c θ + 1} = \frac{1 + s i n θ}{c o s θ}$ , where $θ$ is acute. Verify the result for $θ = 6 0^{\circ}$ .

Try it first, then show the solutionSolution

In the numerator, replace 1 by $sec^{2} θ - tan^{2} θ$ (from $1 + tan^{2} θ = sec^{2} θ$ ): numerator $= (sec θ + tan θ) - (sec^{2} θ - tan^{2} θ)$ . (½ mark)
$= (sec θ + tan θ) - (sec θ + tan θ) (sec θ - tan θ) = (sec θ + tan θ) (1 - sec θ + tan θ)$ . (1 mark)
The denominator $tan θ - sec θ + 1$ is the same as $1 - sec θ + tan θ$ , which is non-zero for an acute angle. Cancel it: LHS $= sec θ + tan θ$ . (½ mark)
$= \frac{1}{c o s θ} + \frac{s i n θ}{c o s θ} = \frac{1 + s i n θ}{c o s θ} =$ RHS. (1 mark)
Check at $θ = 6 0^{\circ}$ : LHS $= \frac{3 + 1}{3 - 1} = \frac{( 3 + 1 ) ^{2}}{2} = \frac{4 + 2 3}{2} = 2 + 3$ . RHS $= \frac{1 + \frac{3}{2}}{\frac{1}{2}} = 2 + 3$ .

Answer: LHS reduces to $sec θ + tan θ = \frac{1 + s i n θ}{c o s θ}$ , so the identity is proved; at $θ = 6 0^{\circ}$ both sides equal $2 + 3$ .

Example 6: Prove the identity, then use it · 5 marks · Long answerHarder board level

(i) In $△ A B C$ , right-angled at $B$ , prove that $sin^{2} A + cos^{2} A = 1$ .
(ii) Hence prove that $1 + tan^{2} A = sec^{2} A$ for $0^{\circ} \leq A < 9 0^{\circ}$ .
(iii) If $3 sin A + 4 cos A = 5$ , where $A$ is acute, find $tan A$ and $sin A$ . (Use the result of (i) on $(3 sin A + 4 cos A)^{2} + (4 sin A - 3 cos A)^{2}$ .)

Try it first, then show the solutionSolution

(i) By Pythagoras theorem, $A B^{2} + B C^{2} = A C^{2}$ . (½ mark)
Divide both sides by $A C^{2}$ : $(\frac{A B}{A C})^{2} + (\frac{B C}{A C})^{2} = 1$ . (½ mark)
Since $cos A = \frac{A B}{A C}$ and $sin A = \frac{B C}{A C}$ , we get $cos^{2} A + sin^{2} A = 1$ . It also holds at $0^{\circ}$ and $9 0^{\circ}$ by direct check: $0 + 1 = 1$ and $1 + 0 = 1$ . (1 mark)
(ii) For $A < 9 0^{\circ}$ , $cos A \neq = 0$ . Divide $sin^{2} A + cos^{2} A = 1$ by $cos^{2} A$ : $tan^{2} A + 1 = sec^{2} A$ . (1 mark)
(iii) $(3 sin A + 4 cos A)^{2} + (4 sin A - 3 cos A)^{2} = 9 sin^{2} A + 24 sin A cos A + 16 cos^{2} A + 16 sin^{2} A - 24 sin A cos A + 9 cos^{2} A = 25 (sin^{2} A + cos^{2} A) = 25$ (1 mark)
Since $3 sin A + 4 cos A = 5$ , we get $25 + (4 sin A - 3 cos A)^{2} = 25$ , so $4 sin A = 3 cos A$ and $tan A = \frac{3}{4}$ . (½ mark)
$sec^{2} A = 1 + \frac{9}{16} = \frac{25}{16}$ , so $cos A = \frac{4}{5}$ and $sin A = 1 - \frac{16}{25} = \frac{3}{5}$ . Check: $3 \cdot \frac{3}{5} + 4 \cdot \frac{4}{5} = \frac{25}{5} = 5$ . (½ mark)

Answer: (i) and (ii) are proved. (iii) $tan A = \frac{3}{4}$ and $sin A = \frac{3}{5}$ .

Example 7: Find the error: square roots of a sum · 3 marks · Short answerHOTS / competency

Aman writes: "Since $sin^{2} A + cos^{2} A = 1$ , taking square roots gives $sin A + cos A = 1$ for every angle $A$ from $0^{\circ}$ to $9 0^{\circ}$ ."
(i) Test his claim at $A = 0^{\circ}$ and at $A = 6 0^{\circ}$ .
(ii) Explain his error, and prove that $sin A + cos A > 1$ for every acute angle $A$ .

Try it first, then show the solutionSolution

At $A = 0^{\circ}$ : $sin 0^{\circ} + cos 0^{\circ} = 0 + 1 = 1$ , so the claim holds here. (½ mark)
At $A = 6 0^{\circ}$ : $\frac{3}{2} + \frac{1}{2} = \frac{3 + 1}{2} > 1$ because $3 > 1$ . So the claim fails. (½ mark)
The error: a square root does not split over a sum, so $p^{2} + q^{2} \neq = p + q$ in general. For example, $9 + 16 = 5$ but $3 + 4 = 7$ . (1 mark)
$(sin A + cos A)^{2} = sin^{2} A + cos^{2} A + 2 sin A cos A = 1 + 2 sin A cos A$ (½ mark)
For an acute angle, $sin A > 0$ and $cos A > 0$ , so $(sin A + cos A)^{2} > 1$ . Since $sin A + cos A$ is positive, $sin A + cos A > 1$ . (½ mark)

Answer: The claim is true at $0^{\circ}$ but false at $6 0^{\circ}$ , where the sum is $\frac{3 + 1}{2}$ . The error is taking the square root term by term. For every acute angle, $sin A + cos A > 1$ .

Example 8: TV screens of the same shape · 3 marks · Short answerHOTS / competency

A rectangular TV screen has a diagonal of 125 cm. The diagonal makes an angle $θ$ with the longer (bottom) edge, where $cot θ = \frac{4}{3}$ .
(i) Find the width and height of the screen.
(ii) A smaller screen of exactly the same shape has a diagonal of 75 cm. Explain why $tan θ$ is the same for both screens, and find the height of the smaller screen.

Try it first, then show the solutionSolution

The bottom edge, a side edge and the diagonal form a right triangle. $cot θ = \frac{width}{height} = \frac{4}{3}$ , so take width $= 4 k$ and height $= 3 k$ . (½ mark)
Diagonal $= 16 k^{2} + 9 k^{2} = 5 k = 125$ , so $k = 25$ . Width $= 100$ cm and height $= 75$ cm. (1 mark)
Same shape means the angles are equal, so the two right triangles are similar (AA). Corresponding sides are then in the same ratio. A trigonometric ratio depends only on the angle, so $tan θ = \frac{3}{4}$ for both screens. (1 mark)
$sin θ = \frac{3}{5}$ , so the height of the smaller screen $= 75 \times \frac{3}{5} = 45$ cm. (Its width is $75 \times \frac{4}{5} = 60$ cm.) (½ mark)

Answer: (i) Width 100 cm, height 75 cm. (ii) $tan θ = \frac{3}{4}$ for both screens because the triangles are similar. The smaller screen is 45 cm high.

Example 9: A condition with root 2: prove, then reason · 3 marks · Short answerHOTS / competency

For an acute angle $θ$ , it is given that $sin θ + cos θ = 2 cos θ$ .
(i) Prove that $cos θ - sin θ = 2 sin θ$ .
(ii) Find $tan θ$ . Neha says $θ$ must be $4 5^{\circ}$ because $2$ appears. Show that she is wrong.

Try it first, then show the solutionSolution

(i) From the given condition, $sin θ = 2 cos θ - cos θ = (2 - 1) cos θ$ . (½ mark)
Multiply both sides by $(2 + 1)$ : $(2 + 1) sin θ = (2 - 1) (2 + 1) cos θ = (2 - 1) cos θ = cos θ$ . (1 mark)
So $2 sin θ + sin θ = cos θ$ , which gives $cos θ - sin θ = 2 sin θ$ . Hence proved. (½ mark)
(ii) Divide $sin θ = (2 - 1) cos θ$ by $cos θ \neq = 0$ : $tan θ = 2 - 1$ . Check with (i): $tan θ = \frac{1}{2 + 1} = 2 - 1$ . (½ mark)
At $θ = 4 5^{\circ}$ : LHS $= \frac{1}{2} + \frac{1}{2} = 2$ , but RHS $= 2 \times \frac{1}{2} = 1$ . They are not equal (also $tan 4 5^{\circ} = 1 \neq = 2 - 1$ ), so $θ$ is not $4 5^{\circ}$ . (½ mark)

Answer: (i) Proved. (ii) $tan θ = 2 - 1$ ; $θ = 4 5^{\circ}$ does not satisfy the condition, so Neha is wrong.

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Rooftop solar panel in Jaipur

Under a rooftop solar scheme, a family in Jaipur installs a flat solar panel on their flat roof. The side edge of the panel is 2 m long. Its lower end rests on the roof, and its upper end is held up by a vertical steel rod. The panel edge, the rod and the roof form a right triangle, with the right angle at the foot of the rod. Let $θ$ be the angle between the panel and the roof. In summer the installer sets $θ = 3 0^{\circ}$ . In winter, to catch the lower sun, he fits a taller rod of height 1.2 m, keeping the panel edge at 2 m.

(i) · 1 mark
Find the height of the rod in summer.
Answer
Answer: 1 m
The panel edge is the hypotenuse and the rod is opposite $θ$ . $sin 3 0^{\circ} = \frac{rod}{2}$ , so rod $= 2 \times \frac{1}{2} = 1$ m.
(ii) · 1 mark
In summer, how far is the foot of the rod from the lower end of the panel?
Answer
Answer: $3$ m (about 1.73 m)
$cos 3 0^{\circ} = \frac{d}{2}$ , so $d = 2 \times \frac{3}{2} = 3$ m.
(iii) · 2 marks
In winter, find $cos θ$ and $tan θ$ , and the new distance between the lower end of the panel and the foot of the rod.
Answer
Answer: $cos θ = \frac{4}{5}$ , $tan θ = \frac{3}{4}$ , distance 1.6 m
$sin θ = \frac{1.2}{2} = \frac{3}{5}$ (½ mark). $cos θ = 1 - \frac{9}{25} = \frac{4}{5}$ (½ mark). Distance $= 2 \times \frac{4}{5} = 1.6$ m (½ mark). $tan θ = \frac{3/5}{4/5} = \frac{3}{4}$ (½ mark).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

If $5 tan θ = 12$ , find the value of $\frac{s i n θ + c o s θ}{s i n θ - c o s θ}$ . (2 marks)

Step	Marks
Divide numerator and denominator by $cos θ$ ( $cos θ \neq = 0$ ): expression $= \frac{t a n θ + 1}{t a n θ - 1}$	0.5
From the given condition, $tan θ = \frac{12}{5}$	0.5
Substitute: $\frac{\frac{12}{5} + 1}{\frac{12}{5} - 1} = \frac{17/5}{7/5}$	0.5
Final answer $= \frac{17}{7}$	0.5

Prove that $\frac{1 + c o s θ - s i n ^{2} θ}{s i n θ ( 1 + c o s θ )} = cot θ$ , where $θ$ is acute. (3 marks)

Step	Marks
Use $1 - sin^{2} θ = cos^{2} θ$ : numerator $= cos^{2} θ + cos θ$	1
Factorise: numerator $= cos θ (cos θ + 1)$ , so LHS $= \frac{c o s θ ( 1 + c o s θ )}{s i n θ ( 1 + c o s θ )}$	1
Cancel $(1 + cos θ) \neq = 0$ : LHS $= \frac{c o s θ}{s i n θ} = cot θ =$ RHS	1

If $sec θ + tan θ = 3$ , where $θ$ is acute, find the value of $sin θ$ . (3 marks)

Step	Marks
Since $sec^{2} θ - tan^{2} θ = 1$ , we have $(sec θ - tan θ) (sec θ + tan θ) = 1$	0.5
So $sec θ - tan θ = \frac{1}{3}$	0.5
Add and subtract the two equations: $2 sec θ = \frac{10}{3}$ and $2 tan θ = \frac{8}{3}$ , so $sec θ = \frac{5}{3}$ and $tan θ = \frac{4}{3}$	1
$sin θ = \frac{t a n θ}{s e c θ} = \frac{4}{3} \times \frac{3}{5} = \frac{4}{5}$ (check: $\frac{5}{3} + \frac{4}{3} = 3$ )	1

If $cosec θ + cot θ = 3$ , where $θ$ is acute, find $sin θ$ and $cos θ$ , and hence evaluate $\frac{2 t a n θ + s e c θ}{c o t θ}$ . (5 marks)

Step	Marks
Since $cosec^{2} θ - cot^{2} θ = 1$ , we have $(cosec θ - cot θ) (cosec θ + cot θ) = 1$	0.5
So $cosec θ - cot θ = \frac{1}{3}$	0.5
Add the two equations: $2 cosec θ = \frac{10}{3}$ , so $cosec θ = \frac{5}{3}$	0.5
Subtract: $2 cot θ = \frac{8}{3}$ , so $cot θ = \frac{4}{3}$	0.5
$sin θ = \frac{1}{cosec θ} = \frac{3}{5}$ and $cos θ = cot θ \times sin θ = \frac{4}{3} \times \frac{3}{5} = \frac{4}{5}$	1
$tan θ = \frac{1}{c o t θ} = \frac{3}{4}$ and $sec θ = \frac{1}{c o s θ} = \frac{5}{4}$	1
Substitute: $\frac{2 \times \frac{3}{4} + \frac{5}{4}}{\frac{4}{3}} = \frac{11}{4} \times \frac{3}{4}$	0.5
Final value $= \frac{33}{16}$	0.5

Where students lose marks in Introduction to Trigonometry

The slips examiners see most often, and what to write instead.

Mistake: Splitting a ratio over a sum of angles
Students treat sin like a number multiplying the bracket, as in algebra.
Wrong working
$sin (3 0^{\circ} + 6 0^{\circ}) = sin 3 0^{\circ} + sin 6 0^{\circ} = \frac{1 + 3}{2}$
Right working
$sin (3 0^{\circ} + 6 0^{\circ}) = sin 9 0^{\circ} = 1$
Do this: Always add the angles first, then take the ratio of the single angle.
Usually costs: The full mark in an MCQ; 1 mark or more in a 2- or 3-mark question.
Mistake: Cancelling the word sin or treating it as a multiplier
Sin is read as a variable instead of a function of the angle.
Wrong working
$\frac{s i n 6 0 ^{\circ}}{s i n 3 0 ^{\circ}} = \frac{60}{30} = 2$
Right working
$\frac{s i n 6 0 ^{\circ}}{s i n 3 0 ^{\circ}} = \frac{3 /2}{1/2} = 3$
Do this: Replace each ratio by its value before doing any arithmetic. $sin A$ is one symbol.
Usually costs: The full mark in an MCQ; usually all marks for that step in longer answers.
Mistake: Squaring only part of a standard value
Rushing, and forgetting that the square applies to the whole fraction.
Wrong working
$cos^{2} 3 0^{\circ} = \frac{3}{2}$ and $tan^{2} 3 0^{\circ} = \frac{1}{9}$
Right working
$cos^{2} 3 0^{\circ} = (\frac{3}{2})^{2} = \frac{3}{4}$ and $tan^{2} 3 0^{\circ} = (\frac{1}{3})^{2} = \frac{1}{3}$
Do this: Write the value in a bracket, then square the numerator and the denominator separately.
Usually costs: ½ to 1 mark, and the final answer is wrong.
Mistake: Using the wrong opposite or adjacent side
Sides are labelled once and reused for the other acute angle.
Wrong working
In $△ A B C$ right-angled at $B$ : $sin C = \frac{B C}{A C}$
Right working
In $△ A B C$ right-angled at $B$ : $sin C = \frac{A B}{A C}$ , because $A B$ is opposite $∠ C$ .
Do this: For each angle, mark it and re-label opposite and adjacent before writing any ratio.
Usually costs: 1 to 2 marks, since every ratio after that is wrong.
Mistake: Treating an identity like an equation
Students cross-multiply or move terms across to reach 1 = 1.
Wrong working
$\frac{1 - s i n ^{2} A}{c o s A} = cos A \Rightarrow 1 - sin^{2} A = cos^{2} A \Rightarrow 1 = 1$ , hence proved.
Right working
LHS $= \frac{1 - s i n ^{2} A}{c o s A} = \frac{c o s ^{2} A}{c o s A} = cos A =$ RHS.
Do this: Start with one side only and transform it until it becomes the other side.
Usually costs: Often 1 to 2 marks of a 3-mark proof, as the method is not accepted as a proof.
Mistake: Giving a value for an undefined ratio
Students guess 0 or infinity because the denominator is 0.
Wrong working
$tan 9 0^{\circ} = 0$ (or $tan 9 0^{\circ} = \infty$ )
Right working
$tan 9 0^{\circ} = \frac{s i n 9 0 ^{\circ}}{c o s 9 0 ^{\circ}} = \frac{1}{0}$ , which is not defined.
Do this: Write the ratio as a fraction; if the denominator is 0, state 'not defined'.
Usually costs: The full mark in an MCQ or assertion-reason item.

NCERT exercise guide: Introduction to Trigonometry

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 8.1

What it tests: Writing the six trigonometric ratios from the sides of a right triangle. Finding a missing side by Pythagoras theorem. Finding all the ratios when one ratio is given, and evaluating an expression from it. True/false reasoning, such as whether a ratio can take a certain value, or whether the ratios change when the triangle is enlarged.

How to approach it: Draw the right triangle and mark the angle. Write the given ratio as two sides with a common multiplier $k$ , find the third side by Pythagoras theorem, then read off any ratio you need. For reasoning items, remember that the hypotenuse is the longest side, so $sin A$ and $cos A$ lie between 0 and 1 for an acute angle. The ratios depend only on the angle, because triangles with the same angles are similar.

Like the questions in Exercise 8.1 where one ratio is given and an expression must be evaluated

If $24 tan θ = 7$ , where $θ$ is acute, find $sin θ$ and $cos θ$ . Hence find the value of $\frac{s e c θ - t a n θ}{s e c θ + t a n θ}$ .

Show the solutionSolution

$tan θ = \frac{7}{24}$ . Take $△ A B C$ right-angled at $B$ with $∠ A = θ$ , so $B C = 7 k$ and $A B = 24 k$ for some positive $k$ .
By Pythagoras theorem, $A C^{2} = (7 k)^{2} + (24 k)^{2} = 49 k^{2} + 576 k^{2} = 625 k^{2}$ , so $A C = 25 k$ .
$sin θ = \frac{B C}{A C} = \frac{7}{25}$ , $cos θ = \frac{A B}{A C} = \frac{24}{25}$ , and $sec θ = \frac{25}{24}$ .
$\frac{s e c θ - t a n θ}{s e c θ + t a n θ} = \frac{\frac{25}{24} - \frac{7}{24}}{\frac{25}{24} + \frac{7}{24}} = \frac{18}{32} = \frac{9}{16}$
Check by a second method: multiply top and bottom by $sec θ - tan θ$ . The denominator becomes $sec^{2} θ - tan^{2} θ = 1$ , so the value is $(sec θ - tan θ)^{2} = (\frac{18}{24})^{2} = (\frac{3}{4})^{2} = \frac{9}{16}$ . It matches.

Answer: $sin θ = \frac{7}{25}$ , $cos θ = \frac{24}{25}$ , and the value is $\frac{9}{16}$ .

Exercise 8.2

What it tests: Evaluating expressions using the values at $0^{\circ}$ , $3 0^{\circ}$ , $4 5^{\circ}$ , $6 0^{\circ}$ and $9 0^{\circ}$ . Picking the correct value in MCQs. Finding angles $A$ and $B$ from two equations involving $A + B$ and $A - B$ . True/false statements, such as whether $sin (A + B) = sin A + sin B$ , or how $sin θ$ and $cos θ$ change as $θ$ increases.

How to approach it: Write the value table from memory first: the $sin$ column is $0, \frac{1}{2}, \frac{1}{2}, \frac{3}{2}, 1$ and the $cos$ column is the same list in reverse. Substitute each value inside brackets, square carefully (for example $tan^{2} 3 0^{\circ} = \frac{1}{3}$ ), and simplify the numerator and denominator separately before dividing. For angle questions, turn each equation into an angle equation such as $A + B = 6 0^{\circ}$ , then solve the pair of linear equations.

Like the evaluation questions in Exercise 8.2

Evaluate: $\frac{4 s i n ^{2} 6 0 ^{\circ} + 3 t a n ^{2} 3 0 ^{\circ} + 8 s i n 4 5 ^{\circ} c o s 4 5 ^{\circ}}{s e c ^{2} 4 5 ^{\circ} + c o t ^{2} 3 0 ^{\circ}}$

Show the solutionSolution

$4 sin^{2} 6 0^{\circ} = 4 \times (\frac{3}{2})^{2} = 4 \times \frac{3}{4} = 3$
$3 tan^{2} 3 0^{\circ} = 3 \times (\frac{1}{3})^{2} = 3 \times \frac{1}{3} = 1$
$8 sin 4 5^{\circ} cos 4 5^{\circ} = 8 \times \frac{1}{2} \times \frac{1}{2} = 8 \times \frac{1}{2} = 4$
Numerator $= 3 + 1 + 4 = 8$
Denominator $= (2)^{2} + (3)^{2} = 2 + 3 = 5$
Value $= \frac{8}{5}$

Answer: $\frac{8}{5}$

Exercise 8.3

What it tests: Expressing all the ratios in terms of one ratio. Simplifying expressions with $sin^{2} A + cos^{2} A = 1$ and its two partner identities. Choosing the correct simplified value in MCQs. Proving simple identities.

How to approach it: Start from the more complicated side. Usually you convert everything to $sin$ and $cos$ , combine fractions over a common denominator, then use an identity to replace $1 - sin^{2} A$ by $cos^{2} A$ , and so on. Never move terms across the equals sign as you would in an equation. Work on one side until it becomes the other side.

Like the identity proofs in Exercise 8.3

Prove that $\frac{1 - s i n θ}{1 + s i n θ} = (sec θ - tan θ)^{2}$ , where $θ$ is acute.

Show the solutionSolution

Start with the right-hand side, which has more to simplify: RHS $= (\frac{1}{c o s θ} - \frac{s i n θ}{c o s θ})^{2}$
$= \frac{( 1 - s i n θ ) ^{2}}{c o s ^{2} θ}$
$= \frac{( 1 - s i n θ ) ^{2}}{1 - s i n ^{2} θ}$ , since $cos^{2} θ = 1 - sin^{2} θ$
$= \frac{( 1 - s i n θ ) ^{2}}{( 1 - s i n θ ) ( 1 + s i n θ )}$ , using $a^{2} - b^{2} = (a - b) (a + b)$
$= \frac{1 - s i n θ}{1 + s i n θ}$ = LHS, cancelling $1 - sin θ \neq = 0$ because $θ$ is acute.

Answer: RHS simplifies to LHS, so the identity is proved.

Introduction to Trigonometry: previous year questions

Standard: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 6.7 marks per paper; 104 questions in all, counting OR alternatives. In 4 of 4 sample papers.
Basic: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 6.6 marks per paper; 92 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Introduction to Trigonometry

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026 Basic, set 430/1/1, Q10 · 1 mark · MCQ

Which of the following statements is false ?

(A) $tan 4 5^{\circ} = cot 4 5^{\circ}$
(B) $sin 9 0^{\circ} = tan 4 5^{\circ}$
(C) $sin 3 0^{\circ} = cos 3 0^{\circ}$
(D) $sin 4 5^{\circ} = cos 4 5^{\circ}$

Try it first, then show our solutionSolution

$tan 4 5^{\circ} = cot 4 5^{\circ} = 1$ , $sin 9 0^{\circ} = tan 4 5^{\circ} = 1$ , and $sin 4 5^{\circ} = cos 4 5^{\circ} = \frac{1}{2}$ , so (A), (B) and (D) are true.
$sin 3 0^{\circ} = \frac{1}{2}$ but $cos 3 0^{\circ} = \frac{3}{2}$ , so statement (C) is false. The answer is option (C).
1 mark

Answer: Option (C)

Where the marks are

Identifies sin 30° ≠ cos 30° (1 mark)

See it in the full 2026 board paper, set 430/1/1

2026 Basic, set 430/1/1, Q24(a) · 2 marks · Very short answer

Find the values of A and B $(0^{\circ} \leq A < 9 0^{\circ}, 0^{\circ} \leq B < 9 0^{\circ})$ , if $tan (A + B) = 1$ and $tan (A - B) = \frac{1}{3}$ .

Try it first, then show our solutionSolution

$tan (A + B) = 1 = tan 4 5^{\circ} \Rightarrow A + B = 4 5^{\circ}$ …(1)
½ mark
$tan (A - B) = \frac{1}{3} = tan 3 0^{\circ} \Rightarrow A - B = 3 0^{\circ}$ …(2)
½ mark
Adding (1) and (2): $2 A = 7 5^{\circ} \Rightarrow A = 37. 5^{\circ}$ .
½ mark
Subtracting (2) from (1): $2 B = 1 5^{\circ} \Rightarrow B = 7. 5^{\circ}$ .
½ mark

Answer: $A = 37. 5^{\circ}$ , $B = 7. 5^{\circ}$

Where the marks are

Gets $A + B = 4 5^{\circ}$ (½ mark)
Gets $A - B = 3 0^{\circ}$ (½ mark)
Finds $A = 37. 5^{\circ}$ (½ mark)
Finds $B = 7. 5^{\circ}$ (½ mark)

See it in the full 2026 board paper, set 430/1/1

2026 Basic, set 430/1/1, Q30 · 3 marks · Short answer

Prove that : $\frac{1 + c o t ^{2} A}{1 + t a n ^{2} A} = (\frac{1 - c o t A}{1 - t a n A})^{2}$

Try it first, then show our solutionSolution

LHS $= \frac{1 + \frac{c o s ^{2} A}{s i n ^{2} A}}{1 + \frac{s i n ^{2} A}{c o s ^{2} A}} = \frac{\frac{s i n ^{2} A + c o s ^{2} A}{s i n ^{2} A}}{\frac{c o s ^{2} A + s i n ^{2} A}{c o s ^{2} A}} = \frac{cos ^{2} A}{sin ^{2} A} = cot^{2} A$ [using $sin^{2} A + cos^{2} A = 1$ ]
1 mark
RHS: $\frac{1 - cot A}{1 - tan A} = \frac{\frac{s i n A - c o s A}{s i n A}}{\frac{c o s A - s i n A}{c o s A}} = \frac{sin A - cos A}{sin A} \times \frac{cos A}{- ( sin A - cos A )} = - \frac{cos A}{sin A} = - cot A$
1½ marks
So RHS $= (- cot A)^{2} = cot^{2} A =$ LHS. Hence proved.
½ mark

Answer: Both sides equal cot²A; hence proved

Where the marks are

Simplifies LHS to $cot^{2} A$ (1 mark)
Simplifies $\frac{1 - c o t A}{1 - t a n A}$ to $- cot A$ (1½ marks)
Concludes LHS = RHS (½ mark)

See it in the full 2026 board paper, set 430/1/1

Every Introduction to Trigonometry question, by year

Standard (123 questions, sample papers included)

For 2027

Sample paper 2026-27: Q8 (1 mark, solved), Q23 (2 marks, solved), Q31(A) (3 marks, solved), Q31(B) (3 marks, solved)

2026

Set 30/1/1: Q9 (1 mark, solved), Q10 (1 mark, solved), Q24(A) (2 marks, solved), Q24(B) (2 marks, solved), Q28(A) (3 marks, solved), Q28(B) (3 marks, solved)

Set 30/2/1: Q10 (1 mark, solved), Q11 (1 mark, solved), Q24(a) (2 marks, solved), Q24(b) (2 marks, solved), Q28(a) (3 marks, solved), Q28(b) (3 marks, solved)

Set 30/3/1: Q15 (1 mark, solved), Q18 (1 mark, solved), Q20 (1 mark, solved), Q26(a) (3 marks, solved), Q26(b) (3 marks, solved)

Set 30/4/1: Q11 (1 mark, solved), Q12 (1 mark, solved), Q23(a) (2 marks, solved), Q23(b) (2 marks, solved), Q28 (3 marks, solved)

Set 30/5/1: Q14 (1 mark, solved), Q18 (1 mark, solved), Q24(a) (2 marks, solved), Q24(b) (2 marks, solved), Q28 (3 marks, solved)

Sample paper 2025-26: Q5 (1 mark, solved), Q9 (1 mark, solved), Q20 (1 mark, solved), Q22 (2 marks, solved), Q29(A) (3 marks, solved), Q29(B) (3 marks, solved)

2025

Set 30/1/1: Q4 (1 mark, solved), Q5 (1 mark, solved), Q21(a) (2 marks, solved), Q21(b) (2 marks, solved), Q27(a) (3 marks, solved), Q27(b) (3 marks, solved)

Set 30/2/1: Q1 (1 mark, solved), Q13 (1 mark, solved), Q23 (2 marks, solved), Q27(a) (3 marks, solved), Q27(b) (3 marks, solved)

Set 30/3/1: Q1 (1 mark, solved), Q9 (1 mark, solved), Q21 (2 marks, solved), Q28(a) (3 marks, solved), Q28(b) (3 marks, solved)

Set 30/4/1: Q8 (1 mark, solved), Q9 (1 mark, solved), Q24(A) (2 marks, solved), Q24(B) (2 marks, solved), Q29(A) (3 marks, solved), Q29(B) (3 marks, solved)

Set 30/5/1: Q12 (1 mark, solved), Q19 (1 mark, solved), Q23(a) (2 marks, solved), Q23(b) (2 marks, solved), Q30(a) (3 marks, solved), Q30(b) (3 marks, solved)

Set 30/6/1: Q12 (1 mark, solved), Q18 (1 mark, solved), Q22(a) (2 marks, solved), Q22(b) (2 marks, solved), Q30(a) (3 marks, solved), Q30(b) (3 marks, solved)

Sample paper 2024-25: Q6 (1 mark, solved), Q12 (1 mark, solved), Q23 (2 marks, solved), Q29 (3 marks, solved)

2024

Set 30/1/1: Q7 (1 mark, solved), Q12 (1 mark, solved), Q23(A) (2 marks, solved), Q23(B) (2 marks, solved), Q28 (3 marks, solved)

Set 30/2/1: Q7 (1 mark, solved), Q19 (1 mark, solved), Q23(a) (2 marks, solved), Q23(b) (2 marks, solved), Q29 (3 marks, solved)

Set 30/3/1: Q8 (1 mark, solved), Q25 (2 marks, solved), Q30 (3 marks, solved)

Set 30/4/1: Q10 (1 mark, solved), Q23(a) (2 marks, solved), Q23(b) (2 marks, solved), Q26 (3 marks, solved)

Set 30/5/1: Q2 (1 mark, solved), Q3 (1 mark, solved), Q22 (2 marks, solved), Q29 (3 marks, solved)

Sample paper 2023-24: Q11 (1 mark, solved), Q12 (1 mark, solved), Q24(a) (2 marks, solved), Q24(b) (2 marks, solved), Q30 (3 marks, solved)

2023

Set 30/1/1: Q6 (1 mark, solved), Q9 (1 mark, solved), Q23(a) (2 marks, solved), Q23(b) (2 marks, solved), Q30 (3 marks, solved)

Set 30/2/1: Q22(a) (2 marks, solved), Q22(b) (2 marks, solved), Q27 (3 marks, solved)

Set 30/4/1: Q10 (1 mark, solved), Q25(A) (2 marks, solved), Q25(B) (2 marks, solved), Q29(A) (3 marks, solved), Q29(B) (3 marks, solved)

Set 30/5/1: Q6 (1 mark, solved), Q12 (1 mark, solved), Q25(a) (2 marks, solved), Q25(b) (2 marks, solved), Q30 (3 marks, solved)

Set 30/6/1: Q7 (1 mark, solved), Q16 (1 mark, solved), Q20 (1 mark, solved), Q30 (3 marks, solved)

Basic (109 questions, sample papers included)

For 2027

Sample paper 2026-27: Q14 (1 mark, solved), Q22 (2 marks, solved), Q29 (3 marks, solved)

2026

Set 430/1/1: Q10 (1 mark, solved), Q11 (1 mark, solved), Q24(a) (2 marks, solved), Q24(b) (2 marks, solved), Q30 (3 marks, solved)

Set 430/2/1: Q10 (1 mark, solved), Q11 (1 mark, solved), Q24 (2 marks, solved), Q30 (3 marks, solved)

Set 430/3/1: Q8 (1 mark, solved), Q19 (1 mark, solved), Q24(a) (2 marks, solved), Q24(b) (2 marks, solved), Q29 (3 marks, solved)

Set 430/4/1: Q13 (1 mark, solved), Q16 (1 mark, solved), Q21 (2 marks, solved), Q28(a) (3 marks, solved), Q28(b) (3 marks, solved)

Set 430/5/1: Q6 (1 mark, solved), Q19 (1 mark, solved), Q21(a) (2 marks, solved), Q21(b) (2 marks, solved), Q29 (3 marks, solved)

Sample paper 2025-26: Q12 (1 mark, solved), Q25(A) (2 marks, solved), Q25(B) (2 marks, solved), Q30 (3 marks, solved)

2025

Set 430/1/1: Q10 (1 mark, solved), Q11 (1 mark, solved), Q24(a) (2 marks, solved), Q24(b) (2 marks, solved), Q30 (3 marks, solved)

Set 430/2/1: Q10 (1 mark, solved), Q11 (1 mark, solved), Q24 (2 marks, solved), Q30 (3 marks, solved)

Set 430/3/1: Q8 (1 mark, solved), Q19 (1 mark, solved), Q24(a) (2 marks, solved), Q24(b) (2 marks, solved), Q29 (3 marks, solved)

Set 430/4/1: Q7 (1 mark, solved), Q11 (1 mark, solved), Q16 (1 mark, solved)

Set 430/5/1: Q12 (1 mark, solved), Q20 (1 mark, solved), Q22(a) (2 marks, solved), Q22(b) (2 marks, solved), Q28 (3 marks, solved)

Set 430/6/1: Q14 (1 mark, solved), Q17 (1 mark, solved), Q24(a) (2 marks, solved), Q30 (3 marks, solved)

Sample paper 2024-25: Q7 (1 mark, solved), Q13 (1 mark, solved), Q24 (2 marks, solved), Q28 (3 marks, solved)

2024

Set 430/1/1: Q4 (1 mark, solved), Q9 (1 mark, solved), Q24 (2 marks, solved), Q31 (3 marks, solved)

Set 430/2/1: Q6 (1 mark, solved), Q7 (1 mark, solved), Q8 (1 mark, solved), Q30 (3 marks, solved)

Set 430/3/1: Q7 (1 mark, solved), Q15 (1 mark, solved), Q22 (2 marks, solved), Q31(a) (3 marks, solved), Q31(b) (3 marks, solved)

Set 430/4/1: Q11 (1 mark, solved), Q25 (2 marks, solved), Q31 (3 marks, solved)

Set 430/5/1: Q10 (1 mark, solved), Q11 (1 mark, solved), Q21 (2 marks, solved), Q28 (3 marks, solved)

Sample paper 2023-24: Q10 (1 mark, solved), Q11 (1 mark, solved), Q12 (1 mark, solved), Q24 (2 marks, solved), Q30(a) (3 marks, solved), Q30(b) (3 marks, solved)

2023

Set 430/1/1: Q2 (1 mark, solved), Q22 (2 marks, solved), Q26(a) (3 marks, solved), Q26(b) (3 marks, solved)

Set 430/2/1: Q3 (1 mark, solved), Q15 (1 mark, solved), Q18 (1 mark, solved), Q29(a) (3 marks, solved), Q29(b) (3 marks, solved)

Set 430/4/1: Q10 (1 mark, solved), Q21 (2 marks, solved), Q27 (3 marks, solved)

Set 430/5/1: Q11 (1 mark, solved), Q16 (1 mark, solved), Q22(a) (2 marks, solved), Q22(b) (2 marks, solved), Q30 (3 marks, solved)

Set 430/6/1: Q3 (1 mark, solved), Q20 (1 mark, solved), Q21 (2 marks, solved), Q31(a) (3 marks, solved), Q31(b) (3 marks, solved)

Practice questions on Introduction to Trigonometry

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
If $sin θ = \frac{1}{3}$ , where $θ$ is acute, then $cot θ$ equals
(a) $22$
(b) $\frac{1}{2 2}$
(c) $\frac{3}{2 2}$
(d) $\frac{2 2}{3}$
Hint
Find $cos θ$ from $sin^{2} θ + cos^{2} θ = 1$ first.
Answer
(a) $22$
$cos θ = 1 - \frac{1}{9} = \frac{2 2}{3}$ , so $cot θ = \frac{c o s θ}{s i n θ} = \frac{2 2 /3}{1/3} = 22$ .
Q2 · 2 marks · Very short answerBoard level
$△ A B C$ is right-angled at $B$ and $△ P QR$ is right-angled at $Q$ , with $∠ A = ∠ P$ . Show that $sin A = sin P$ and $tan A = tan P$ . What does this tell you about trigonometric ratios?
Hint
Two pairs of equal angles give similar triangles.
Answer
$∠ B = ∠ Q = 9 0^{\circ}$ and $∠ A = ∠ P$ , so $△ A B C \sim △ P QR$ (AA). Hence $\frac{B C}{QR} = \frac{A C}{P R} = \frac{A B}{P Q}$ .
So $\frac{B C}{A C} = \frac{QR}{P R}$ , i.e. $sin A = sin P$ , and $\frac{B C}{A B} = \frac{QR}{P Q}$ , i.e. $tan A = tan P$ .
The ratios depend only on the angle, not on the size of the triangle, so they are well defined.
Q3 · 2 marks · Very short answerBoard level
Which of $sec 9 0^{\circ}$ , $cosec 9 0^{\circ}$ and $cot 9 0^{\circ}$ are defined? Give a reason, and find the sum of the ones that are defined.
Hint
Write each ratio in terms of $sin 9 0^{\circ} = 1$ and $cos 9 0^{\circ} = 0$ .
Answer
$sec 9 0^{\circ} = \frac{1}{c o s 9 0 ^{\circ}} = \frac{1}{0}$ , which is not defined.
$cosec 9 0^{\circ} = \frac{1}{1} = 1$ and $cot 9 0^{\circ} = \frac{0}{1} = 0$ are defined.
Sum $= 1 + 0 = 1$ .
Q4 · 3 marks · Short answerBoard level
If $sin θ + sin^{2} θ = 1$ , prove that $cos^{2} θ + cos^{4} θ = 1$ .
Hint
Rewrite the given condition as $sin θ = 1 - sin^{2} θ = cos^{2} θ$ .
Answer
From the given condition, $sin θ = 1 - sin^{2} θ = cos^{2} θ$ .
Then $cos^{4} θ = (cos^{2} θ)^{2} = sin^{2} θ$ .
So $cos^{2} θ + cos^{4} θ = sin θ + sin^{2} θ = 1$ , as given.
Q5 · 3 marks · Short answerBoard level
Prove that $\frac{t a n A}{s e c A - 1} + \frac{t a n A}{s e c A + 1} = 2 cosec A$ , where $A$ is acute.
Hint
Take the LCM; the denominator becomes $sec^{2} A - 1$ .
Answer
LHS $= \frac{t a n A [( s e c A + 1 ) + ( s e c A - 1 )]}{s e c ^{2} A - 1} = \frac{2 s e c A t a n A}{t a n ^{2} A} = \frac{2 s e c A}{t a n A}$ .
$= \frac{2}{c o s A} \times \frac{c o s A}{s i n A} = \frac{2}{s i n A} = 2 cosec A$ = RHS.
Q6 · 3 marks · Short answerBoard level
Find $x$ if $\frac{x cosec ^{2} 3 0 ^{\circ} s e c ^{2} 4 5 ^{\circ}}{8 c o s ^{2} 4 5 ^{\circ} s i n ^{2} 6 0 ^{\circ}} = tan^{2} 6 0^{\circ} - tan^{2} 3 0^{\circ}$ .
Hint
Simplify each side to a number first, then solve the linear equation.
Answer
LHS $= \frac{x \times 4 \times 2}{8 \times \frac{1}{2} \times \frac{3}{4}} = \frac{8 x}{3}$ .
RHS $= 3 - \frac{1}{3} = \frac{8}{3}$ .
So $\frac{8 x}{3} = \frac{8}{3}$ , which gives $x = 1$ .
Q7 · 5 marks · Long answerHarder board level
In $△ A B C$ , right-angled at $C$ , $∠ A = 3 0^{\circ}$ and $A B = 20$ cm. $D$ is a point on $A C$ such that $∠ D B C = 4 5^{\circ}$ . Find (i) $B C$ and $A C$ , (ii) $C D$ and $B D$ , (iii) the perimeter of $△ A B D$ in simplest surd form.
Hint
$A B$ is the hypotenuse of $△ A B C$ . In $△ D B C$ , the side $B C$ is adjacent to the $4 5^{\circ}$ angle.
Answer
(i) $B C = A B sin 3 0^{\circ} = 10$ cm and $A C = A B cos 3 0^{\circ} = 103$ cm.
(ii) In $△ D B C$ , $tan 4 5^{\circ} = \frac{C D}{B C}$ , so $C D = 10$ cm, and $B D = \frac{B C}{c o s 4 5 ^{\circ}} = 102$ cm.
(iii) $A D = 103 - 10$ cm. Perimeter $= 20 + 102 + 103 - 10 = 10 (1 + 2 + 3)$ cm.
Q8 · 5 marks · Long answerHarder board level
If $cosec θ + cot θ = 3$ , where $θ$ is acute, (i) find $cosec θ - cot θ$ , (ii) find $sin θ$ and $cos θ$ , (iii) hence find the value of $\frac{2 t a n θ + s e c θ}{c o t θ}$ .
Hint
Use $(cosec θ - cot θ) (cosec θ + cot θ) = 1$ , which comes from $1 + cot^{2} θ = cosec^{2} θ$ .
Answer
(i) $cosec θ - cot θ = \frac{1}{3}$ .
(ii) Adding the two equations: $2 cosec θ = \frac{10}{3}$ , so $cosec θ = \frac{5}{3}$ and $cot θ = \frac{4}{3}$ . Hence $sin θ = \frac{3}{5}$ and $cos θ = cot θ \times sin θ = \frac{4}{5}$ .
(iii) $tan θ = \frac{3}{4}$ and $sec θ = \frac{5}{4}$ , so the value is $\frac{\frac{6}{4} + \frac{5}{4}}{\frac{4}{3}} = \frac{11}{4} \times \frac{3}{4} = \frac{33}{16}$ .
Q9 · 5 marks · Long answerHOTS / competency
(i) For an acute angle $θ$ , prove that $sin θ cos θ \leq \frac{1}{2}$ .
(ii) If $sin θ + cos θ = 2$ , where $θ$ is acute, find $sin θ cos θ$ , the value of $tan θ + cot θ$ , and the angle $θ$ .
(iii) Riya says that $sin θ + cos θ = \frac{3}{2}$ for some acute angle $θ$ . Use (i) to show that she is wrong.
Hint
Expand $(sin θ - cos θ)^{2} \geq 0$ and $(sin θ + cos θ)^{2}$ , and use $sin^{2} θ + cos^{2} θ = 1$ .
Answer
(i) $(sin θ - cos θ)^{2} \geq 0$ , so $sin^{2} θ + cos^{2} θ - 2 sin θ cos θ \geq 0$ , i.e. $1 - 2 sin θ cos θ \geq 0$ . Hence $sin θ cos θ \leq \frac{1}{2}$ . (1 mark)
(ii) Squaring: $sin^{2} θ + cos^{2} θ + 2 sin θ cos θ = 2$ , so $1 + 2 sin θ cos θ = 2$ and $sin θ cos θ = \frac{1}{2}$ . (1 mark)
$tan θ + cot θ = \frac{s i n ^{2} θ + c o s ^{2} θ}{s i n θ c o s θ} = \frac{1}{1/2} = 2$ . (1 mark)
$(sin θ - cos θ)^{2} = 1 - 2 \times \frac{1}{2} = 0$ , so $sin θ = cos θ$ , $tan θ = 1$ and $θ = 4 5^{\circ}$ . Check: $sin 4 5^{\circ} + cos 4 5^{\circ} = \frac{2}{2} = 2$ . (1 mark)
(iii) If $sin θ + cos θ = \frac{3}{2}$ , squaring gives $1 + 2 sin θ cos θ = \frac{9}{4}$ , so $sin θ cos θ = \frac{5}{8}$ . But $\frac{5}{8} > \frac{1}{2}$ , which contradicts (i). So no such acute angle exists. (1 mark)
Q10 · 3 marks · Short answerHOTS / competency
A town's building guideline says a loading ramp is unsafe if it makes an angle of more than $3 0^{\circ}$ with the ground. Ramp X is 10 m long along its slope, and its top is 5 m above the ground. Ramp Y covers a horizontal distance of 4 m, and its top is 4 m above the ground.
(i) Find the angle each ramp makes with the ground, and say which ramp is unsafe.
(ii) Keeping the horizontal distance of Y at 4 m, to what height must its top be lowered so that it makes exactly $3 0^{\circ}$ with the ground?
Hint
For X you know the hypotenuse, so use sin. For Y you know both legs, so use tan.
Answer
(i) Ramp X: $sin θ = \frac{5}{10} = \frac{1}{2}$ , so $θ = 3 0^{\circ}$ , which is allowed. Ramp Y: $tan θ = \frac{4}{4} = 1$ , so $θ = 4 5^{\circ} > 3 0^{\circ}$ . Ramp Y is unsafe.
(ii) $tan 3 0^{\circ} = \frac{h}{4}$ , so $h = \frac{4}{3} = \frac{4 3}{3}$ m (about 2.31 m).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.In $△ A B C$ right-angled at $B$ , what is $cos A$ in terms of sides?Show answer
$cos A = \frac{A B}{A C}$ (adjacent over hypotenuse).
2.Why does $sin A$ not depend on the size of the triangle?Show answer
All right triangles with the same acute angle $A$ are similar (AA), so their corresponding sides are proportional and the ratio stays the same.
3.Value of $tan 6 0^{\circ}$ ?Show answer
$3$
4.Which ratios are not defined at $0^{\circ}$ ?Show answer
$cosec 0^{\circ}$ and $cot 0^{\circ}$ , because $sin 0^{\circ} = 0$ .
5.Which ratios are not defined at $9 0^{\circ}$ ?Show answer
$tan 9 0^{\circ}$ and $sec 9 0^{\circ}$ , because $cos 9 0^{\circ} = 0$ .
6.Complete: $1 + tan^{2} A = ?$ Show answer
$sec^{2} A$ , valid for $0^{\circ} \leq A < 9 0^{\circ}$ .
7.If $sec θ + tan θ = 3$ , what is $sec θ - tan θ$ ?Show answer
$\frac{1}{3}$ , since their product is $sec^{2} θ - tan^{2} θ = 1$ .
8.Can $sin A = \frac{5}{4}$ for an acute angle $A$ ?Show answer
No. The opposite side is shorter than the hypotenuse, so $sin A < 1$ .

Best order to study it

Revise Pythagoras theorem and AA similarity, then learn the six ratios and practise labelling opposite and adjacent for each acute angle.
Understand why the ratios are well defined, and practise finding all ratios from one given ratio using sides $ak$ , $bk$ .
Derive the values at $3 0^{\circ}$ , $4 5^{\circ}$ , $6 0^{\circ}$ from the two special triangles, then learn the values at $0^{\circ}$ and $9 0^{\circ}$ and which ratios are not defined.
Practise evaluating expressions and finding unknown angles from equations using the table.
Learn the reciprocal and quotient relations, prove $sin^{2} A + cos^{2} A = 1$ and its two partners, and practise simple identity proofs, including ones with a given condition.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Write all six ratios of $∠ A$ and of $∠ C$ for $△ A B C$ right-angled at $B$ .
Explain in two lines, using similar triangles, why $tan A$ depends only on $A$ .
Write the full table for $0^{\circ}, 3 0^{\circ}, 4 5^{\circ}, 6 0^{\circ}, 9 0^{\circ}$ , including every 'not defined'.
Given $cos θ = \frac{12}{13}$ , find $tan θ$ and $cosec θ$ .
Evaluate $4 sin^{2} 6 0^{\circ} + 3 tan^{2} 3 0^{\circ} - 8 sin 4 5^{\circ} cos 4 5^{\circ}$ (answer: 0).
Solve $2 sin 3 x = 3$ for acute $3 x$ (answer: $x = 2 0^{\circ}$ ).
Prove $sin^{2} A + cos^{2} A = 1$ from Pythagoras, then derive the other two identities.
If $cosec θ + cot θ = k$ , write $cosec θ - cot θ$ (answer: $\frac{1}{k}$ ).
Prove $\frac{1 + t a n A}{1 + c o t A} = tan A$ without looking.
State the range of $sin A$ , $cos A$ and $sec A$ for an acute angle.

Introduction to Trigonometry: frequently asked questions

Are trigonometric ratios of complementary angles in the 2026-27 syllabus?

No. Trigonometric ratios of complementary angles were removed in the NCERT rationalisation and are not in the 2026-27 syllabus.

Should I memorise the table of values or derive it?

Do both. Memorise the table so you are fast in MCQs, but know how it comes from the isosceles right triangle and the half equilateral triangle, so you can rebuild any value you forget.

How do I start proving an identity when I feel stuck?

Pick the more complicated side and convert every ratio into sin and cos. Take the LCM, look for sin squared plus cos squared, and factorise. If one side still will not simplify, work on the other side separately until both match.

Why is tan 90 degrees not defined?

Tan is sin divided by cos, and cos 90 degrees is 0. Division by zero has no meaning, so tan 90 degrees is not defined. The same reason makes sec 90 degrees, cot 0 degrees and cosec 0 degrees undefined.

How is this chapter connected to heights and distances?

The next chapter, Some Applications of Trigonometry, uses exactly these ratios and standard values to find heights of towers and distances between objects. If you can label a right triangle and recall the values at 30, 45 and 60 degrees confidently, that chapter becomes much easier.

Based on: CBSE Mathematics syllabus 2026-27; NCERT Mathematics, Textbook for Class X (Reprint 2026-27)