MathsCBSE Class 10 · Introduction to Trigonometry

Proving Simple Trigonometric Identities for Class 10

Updated 14 October 2026 · Fact-checked

Proving a trigonometric identity means showing that two expressions are equal for every allowed angle. You start from one side and change it step by step into the other, using the reciprocal and quotient relations and sin squared plus cos squared equals one. The board asks two- and three-mark proofs, verification questions, and identities with a given condition.

Start from one side, rewrite it in sin and cos, use sin²A + cos²A = 1, and reach the other side; never cross-multiply.

Illustration drawn for QuestionBuddy.

Understand Proving Simple Trigonometric Identities

An identity is an equation that is true for every angle at which both sides are defined. For example, $sin^{2} A + cos^{2} A = 1$ holds for every $A$ from $0^{\circ}$ to $9 0^{\circ}$ . In contrast, $2 sin A = 1$ is an equation that is true only when $A = 3 0^{\circ}$ . When you prove an identity, you are not solving for $A$ . You are showing that the two sides are the same expression written in two different ways.

Because you are trying to show that the two sides are equal, you cannot use their equality as a starting point. So you must not cross-multiply or move terms across the $=$ sign. Instead, take one side, usually the more complicated one, and rewrite it one valid step at a time until it becomes the other side. Every step must be either algebra (LCM, expanding, factorising, cancelling a non-zero factor) or a known result.

The tools come from the rest of the chapter. The reciprocal relations $cosec A = \frac{1}{s i n A}$ , $sec A = \frac{1}{c o s A}$ and $cot A = \frac{1}{t a n A}$ cut six ratios down to three. The quotient relations $tan A = \frac{s i n A}{c o s A}$ and $cot A = \frac{c o s A}{s i n A}$ cut them down to two. The fundamental identity $sin^{2} A + cos^{2} A = 1$ and its partners $sec^{2} A - tan^{2} A = 1$ and $cosec^{2} A - cot^{2} A = 1$ let you swap a squared expression for a simpler one. The partners come straight from the first identity. Divide $sin^{2} A + cos^{2} A = 1$ by $cos^{2} A$ (allowed when $A \neq = 9 0^{\circ}$ ) to get $tan^{2} A + 1 = sec^{2} A$ . Divide it by $sin^{2} A$ (allowed when $A \neq = 0^{\circ}$ ) to get $1 + cot^{2} A = cosec^{2} A$ . That is why each partner holds wherever its ratios are defined.

The most useful habit is to write everything in $sin$ and $cos$ . Then most proofs become fraction algebra. Watch for $1 - sin^{2} A$ (which is $cos^{2} A$ ), for $a^{2} - b^{2} = (a - b) (a + b)$ , and for the pair $(sec A - tan A) (sec A + tan A) = 1$ . Multiplying top and bottom by a conjugate such as $1 + sin A$ often opens up a stuck proof.

Conditions matter. An identity holds only where every term is defined and you never divide by zero. So $tan A$ and $sec A$ need $A \neq = 9 0^{\circ}$ , $cot A$ and $cosec A$ need $A \neq = 0^{\circ}$ , and if you cancel a factor like $cos A - sin A$ you need $A \neq = 4 5^{\circ}$ . Board questions usually say " $A$ is acute". When you cancel a factor that could be zero, state that it is non-zero. Square roots also need care. For an acute angle every ratio is positive, so $cos^{2} A = cos A$ .

Proving Simple Trigonometric Identities: formulas to remember

Fundamental identity: $sin^{2} A + cos^{2} A = 1$; Valid for 0° ≤ A ≤ 90°. Rearranged forms: cos²A = 1 − sin²A and sin²A = 1 − cos²A.
Sec–tan identity: $sec^{2} A - tan^{2} A = 1$; Valid for 0° ≤ A < 90°. Same as 1 + tan²A = sec²A.
Cosec–cot identity: $cosec^{2} A - cot^{2} A = 1$; Valid for 0° < A ≤ 90°. Same as 1 + cot²A = cosec²A.
Factorised sec–tan form: $(sec A - tan A) (sec A + tan A) = 1$; If sec A + tan A = p, then sec A − tan A = 1/p.
Factorised cosec–cot form: $(cosec A - cot A) (cosec A + cot A) = 1$; If cosec A − cot A = q, then cosec A + cot A = 1/q.
Conjugate forms: $(1 - sin A) (1 + sin A) = cos^{2} A, (1 - cos A) (1 + cos A) = sin^{2} A$; Multiply top and bottom by the conjugate to clear 1 ± sin A or 1 ± cos A from a denominator.
Reciprocal and quotient relations: $tan A = \frac{s i n A}{c o s A}, cot A = \frac{c o s A}{s i n A}, sec A = \frac{1}{c o s A}, cosec A = \frac{1}{s i n A}$; Convert everything to sin and cos first when you are unsure how to start.
Useful algebra: $a^{2} - b^{2} = (a - b) (a + b), (a + b)^{2} = a^{2} + 2 ab + b^{2}$; Most proofs finish with a factorisation or an expansion.

How to solve Proving Simple Trigonometric Identities questions

This method works for any simple identity in the board paper. Write the proof in a column, one step per line, starting from "LHS =" or "RHS =".

1Read both sides and pick the more complicated one to start from. This is usually the one with fractions, squares or more terms.
2If you see no obvious identity, convert every $tan$ , $cot$ , $sec$ and $cosec$ into $sin$ and $cos$ .
3Combine fractions using the LCM. Simplify the numerator and the denominator separately.
4Look for $1 - sin^{2} A$ , $1 - cos^{2} A$ , $sec^{2} A - tan^{2} A$ or $cosec^{2} A - cot^{2} A$ and replace each with its simpler form.
5Factorise, especially a difference of squares, and cancel common factors. State that the factor you cancel is non-zero.
6If a denominator is $1 \pm sin A$ or $1 \pm cos A$ , multiply the top and bottom by its conjugate.
7If you are stuck, simplify the other side separately and show that both sides reach the same expression. End with "= RHS. Hence proved."
8If the question asks you to verify the result, substitute a standard angle such as $3 0^{\circ}$ or $6 0^{\circ}$ into both sides.

Proving Simple Trigonometric Identities in real life

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.

Proving simple trigonometric identities

Suppose you are flying a kite at a Makar Sankranti gathering on flat open ground. The string is pulled straight, and the kite is directly above a point 30 m away from you along the ground. The string makes an angle $θ$ with the ground, and a measuring tool shows only that $sec θ + tan θ = 3$ .

Using an identity, find $sec θ$ and $tan θ$ . Hence find the length of the string and the height of the kite above the ground. Treat the string as a straight line from your hand at ground level.

Try it first, then show the solutionSolution

The string, the ground and the vertical below the kite form a right triangle. The ground is the side adjacent to $θ$ and is 30 m long.
Use the identity $(sec θ - tan θ) (sec θ + tan θ) = sec^{2} θ - tan^{2} θ = 1$ .
So $sec θ - tan θ = \frac{1}{3}$ .
Add the two equations: $2 sec θ = 3 + \frac{1}{3} = \frac{10}{3}$ , so $sec θ = \frac{5}{3}$ .
Subtract them: $2 tan θ = 3 - \frac{1}{3} = \frac{8}{3}$ , so $tan θ = \frac{4}{3}$ .
Length of string $= 30 \times sec θ = 30 \times \frac{5}{3} = 50$ m.
Height of kite $= 30 \times tan θ = 30 \times \frac{4}{3} = 40$ m.
Check by Pythagoras: $3 0^{2} + 4 0^{2} = 900 + 1600 = 2500 = 5 0^{2}$ . Correct.

Answer: $sec θ = \frac{5}{3}$ and $tan θ = \frac{4}{3}$ . The string is 50 m long and the kite is 40 m above the ground.

Memory hook: Kite trick: when sec + tan is given, sec − tan is just its reciprocal.

Tricks to remember Proving Simple Trigonometric Identities

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

SC first: Sine-Cosine first
When you are stuck, turn everything into $sin$ and $cos$ . Six ratios become two, and the proof becomes fraction algebra.
Use it when: Use this as the opening move for any identity mixing $tan$ , $sec$ , $cot$ and $cosec$ .
One minus, one square
Whenever you see 1 minus a square, look for its partner: $1 - sin^{2} = cos^{2}$ , $1 - cos^{2} = sin^{2}$ , $sec^{2} - 1 = tan^{2}$ , $cosec^{2} - 1 = cot^{2}$ .
Use it when: Use this in the middle of a proof to collapse two terms into one.
Partners multiply to 1
$sec \pm tan$ are partners, and so are $cosec \pm cot$ . Each pair multiplies to 1, just like $(a - b) (a + b) = a^{2} - b^{2}$ .
Use it when: Use this when a question gives $sec θ + tan θ$ or has $sec A - tan A$ in a denominator.
Never cross the bridge
Never cross-multiply or move terms across the $=$ sign. Simplify one side into the other, or simplify each side separately until they meet.
Use it when: Use this as a reminder of the correct proof format, so that you keep your method marks.

Proving Simple Trigonometric Identities: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Using the sec–tan partners · 2 marks · Very short answerBoard level

Prove that $sec A (1 - sin A) (sec A + tan A) = 1$ , where $A$ is acute.

Try it first, then show the solutionSolution

LHS $= sec A (1 - sin A) (sec A + tan A)$ .
$sec A (1 - sin A) = \frac{1}{c o s A} - \frac{s i n A}{c o s A} = sec A - tan A$ . (1 mark)
So LHS $= (sec A - tan A) (sec A + tan A) = sec^{2} A - tan^{2} A$ .
$= 1$ , using $sec^{2} A - tan^{2} A = 1$ . LHS = RHS. Hence proved. (1 mark)

Answer: LHS simplifies to $sec^{2} A - tan^{2} A = 1$ , which is the RHS.

Example 2: Proof with a square root and a conjugate · 3 marks · Short answerHarder board level

Prove that $\frac{1 + s i n A}{1 - s i n A} = sec A + tan A$ , where $A$ is acute.

Try it first, then show the solutionSolution

LHS $= \frac{1 + s i n A}{1 - s i n A} \times \frac{1 + s i n A}{1 + s i n A}$ . Multiply top and bottom by the conjugate $1 + sin A$ .
$= \frac{( 1 + s i n A ) ^{2}}{1 - s i n ^{2} A} = \frac{( 1 + s i n A ) ^{2}}{c o s ^{2} A}$ , since $1 - sin^{2} A = cos^{2} A$ . (1 mark)
$= \frac{1 + s i n A}{c o s A}$ . Both $1 + sin A$ and $cos A$ are positive for an acute angle, so we take the positive root. (1 mark)
$= \frac{1}{c o s A} + \frac{s i n A}{c o s A} = sec A + tan A =$ RHS. Hence proved. (1 mark)

Answer: LHS $= \frac{1 + s i n A}{c o s A} = sec A + tan A$ = RHS.

Example 3: Factorising cubes down to tan A · 3 marks · Short answerHarder board level

Prove that $\frac{s i n A - 2 s i n ^{3} A}{2 c o s ^{3} A - c o s A} = tan A$ , where $A$ is acute and $A \neq = 4 5^{\circ}$ .

Try it first, then show the solutionSolution

LHS $= \frac{s i n A ( 1 - 2 s i n ^{2} A )}{c o s A ( 2 c o s ^{2} A - 1 )}$ , taking out the common factors. (1 mark)
Now $1 - 2 sin^{2} A = 1 - 2 (1 - cos^{2} A) = 2 cos^{2} A - 1$ , using $sin^{2} A = 1 - cos^{2} A$ . (1 mark)
So LHS $= \frac{s i n A ( 2 c o s ^{2} A - 1 )}{c o s A ( 2 c o s ^{2} A - 1 )}$ .
$2 cos^{2} A - 1 \neq = 0$ because $A \neq = 4 5^{\circ}$ . Cancelling it gives $\frac{s i n A}{c o s A} = tan A =$ RHS. Hence proved. (1 mark)

Answer: After factorising, the numerator and denominator share the factor $2 cos^{2} A - 1$ . Cancelling it leaves $tan A$ .

Example 4: Dividing through by a common ratio · 2 marks · Very short answerBoard level

Prove that $\frac{c o t A - c o s A}{c o t A + c o s A} = \frac{cosec A - 1}{cosec A + 1}$ , where $A$ is acute.

Try it first, then show the solutionSolution

LHS $= \frac{\frac{c o s A}{s i n A} - c o s A}{\frac{c o s A}{s i n A} + c o s A}$ .
Divide the numerator and denominator by $cos A$ , which is non-zero: LHS $= \frac{\frac{1}{s i n A} - 1}{\frac{1}{s i n A} + 1}$ . (1 mark)
$= \frac{cosec A - 1}{cosec A + 1} =$ RHS. Hence proved. (1 mark)

Answer: Dividing the numerator and the denominator by $cos A$ gives $\frac{cosec A - 1}{cosec A + 1}$ .

Example 5: A product that collapses to a number · 3 marks · Short answerHarder board level

Prove that $(1 + cot A - cosec A) (1 + tan A + sec A) = 2$ , where $A$ is acute. Verify the result for $A = 4 5^{\circ}$ .

Try it first, then show the solutionSolution

Write each factor in $sin$ and $cos$ : $1 + cot A - cosec A = \frac{s i n A + c o s A - 1}{s i n A}$ and $1 + tan A + sec A = \frac{c o s A + s i n A + 1}{c o s A}$ . (1 mark)
LHS $= \frac{( s i n A + c o s A ) ^{2} - 1 ^{2}}{s i n A c o s A}$ , using $(a - b) (a + b) = a^{2} - b^{2}$ .
$= \frac{s i n ^{2} A + c o s ^{2} A + 2 s i n A c o s A - 1}{s i n A c o s A} = \frac{2 s i n A c o s A}{s i n A c o s A} = 2 =$ RHS. (1 mark)
Verify at $4 5^{\circ}$ : $(1 + 1 - 2) (1 + 1 + 2) = (2 - 2) (2 + 2) = 4 - 2 = 2$ . Verified. (1 mark)

Answer: The product equals $\frac{( s i n A + c o s A ) ^{2} - 1}{s i n A c o s A} = \frac{2 s i n A c o s A}{s i n A c o s A} = 2$ . At $4 5^{\circ}$ it gives $(2 - 2) (2 + 2) = 2$ .

Example 6: Identity from two given conditions · 3 marks · Short answerHOTS / competency

For an acute angle $θ$ , let $tan θ + sin θ = m$ and $tan θ - sin θ = n$ . Prove that $m^{2} - n^{2} = 4 mn$ .

Try it first, then show the solutionSolution

$m^{2} - n^{2} = (m + n) (m - n) = (2 tan θ) (2 sin θ) = 4 tan θ sin θ$ . (1 mark)
$mn = tan^{2} θ - sin^{2} θ = \frac{s i n ^{2} θ}{c o s ^{2} θ} - sin^{2} θ = sin^{2} θ \cdot \frac{1 - c o s ^{2} θ}{c o s ^{2} θ}$ .
$= sin^{2} θ \cdot \frac{s i n ^{2} θ}{c o s ^{2} θ} = sin^{2} θ tan^{2} θ$ . (1 mark)
So $mn = sin θ tan θ$ , which is positive because $θ$ is acute. Then $4 mn = 4 tan θ sin θ = m^{2} - n^{2}$ . Hence proved. (1 mark)

Answer: Both $m^{2} - n^{2}$ and $4 mn$ equal $4 tan θ sin θ$ , so $m^{2} - n^{2} = 4 mn$ .

Where students lose marks in Proving Simple Trigonometric Identities

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

Mistake: Treating the identity as an equation and cross-multiplying
This assumes the very equality you are asked to prove. Examiners give little or no credit for such a proof.
Wrong working
To prove $\frac{c o s A}{1 - s i n A} = \frac{1 + s i n A}{c o s A}$ : cross-multiply to get $cos^{2} A = 1 - sin^{2} A$ , which is true, so the result is proved.
Right working
LHS $= \frac{c o s A ( 1 + s i n A )}{( 1 - s i n A ) ( 1 + s i n A )} = \frac{c o s A ( 1 + s i n A )}{c o s ^{2} A} = \frac{1 + s i n A}{c o s A} =$ RHS.
Do this: Work on one side only and end with "= RHS".
Usually costs: Most of the marks for the proof.
Mistake: Cancelling terms instead of factors
You can cancel only a factor that multiplies the whole numerator and the whole denominator, never a term that is being added.
Wrong working
$\frac{s i n A + s i n A c o s A}{s i n A} = 1 + sin A cos A$ , after 'cancelling' $sin A$ from only the first term of the numerator.
Right working
$\frac{s i n A + s i n A c o s A}{s i n A} = \frac{s i n A ( 1 + c o s A )}{s i n A} = 1 + cos A$ , since $sin A \neq = 0$ for an acute angle. Factorise first, then cancel.
Do this: Before you cancel anything, factorise the numerator and the denominator completely.
Usually costs: 1 to 2 marks, and every step after the error.
Mistake: Getting the sign of a partner identity wrong
Students mix up $sec^{2} A - tan^{2} A = 1$ with $tan^{2} A - sec^{2} A = 1$ , or write $sec^{2} A + tan^{2} A = 1$ .
Wrong working
$sec^{2} A - 1 = - tan^{2} A$
Right working
$sec^{2} A = 1 + tan^{2} A$ , so $sec^{2} A - 1 = tan^{2} A$ . Similarly, $cosec^{2} A - 1 = cot^{2} A$ .
Do this: Remember that $sec$ and $cosec$ are at least 1, so they are the bigger squares. Subtract the smaller square from them.
Usually costs: 1 mark for the step and the final answer.
Mistake: Squaring incorrectly or ignoring the sign of a root
$(sin A + cos A)^{2}$ is not $sin^{2} A + cos^{2} A$ ; the middle term is missing. Also, the square root needs the sign fixed by the fact that the angle is acute.
Wrong working
$(sin A + cos A)^{2} - 1 = sin^{2} A + cos^{2} A - 1 = 0$
Right working
$(sin A + cos A)^{2} - 1 = 1 + 2 sin A cos A - 1 = 2 sin A cos A$
Do this: Write out $a^{2} + 2 ab + b^{2}$ in full every time. When you take a root, add a note such as "positive, since $A$ is acute".
Usually costs: 1 to 2 marks.

Exam tips

Write each step on its own line, starting from "LHS =" and ending with "= RHS. Hence proved." Step marks are given for each correct use of an identity.
Name the identity you use in brackets, for example "(since $1 - sin^{2} A = cos^{2} A$ )". It shows the examiner your reason for the step.
If a question also says "verify for $A = 6 0^{\circ}$ ", substitute the exact values into both sides. A verification is not a proof, so do both parts.
When you cancel a factor such as $cos A - sin A$ or $2 cos^{2} A - 1$ , state that it is non-zero for the given angle.
Save time: try the $sin$ – $cos$ conversion for about two lines. If it gets messy, try the conjugate or the partner-product method instead.

Practice questions on Proving Simple Trigonometric Identities

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

Q1 · 2 marks · Very short answerBoard level
Prove that $\frac{1 + t a n ^{2} A}{1 + c o t ^{2} A} = tan^{2} A$ , where $A$ is acute.
Hint
Replace the numerator and the denominator using the two partner identities.
Answer
LHS $= \frac{s e c ^{2} A}{cosec ^{2} A} = \frac{1/ c o s ^{2} A}{1/ s i n ^{2} A} = \frac{s i n ^{2} A}{c o s ^{2} A} = tan^{2} A =$ RHS.
Q2 · 3 marks · Short answerHarder board 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 common denominator becomes $sec^{2} A - 1 = tan^{2} A$ .
Answer
LHS $= tan A \cdot \frac{( s e c A + 1 ) + ( s e c A - 1 )}{s e c ^{2} A - 1} = \frac{2 t a n A s e c A}{t a n ^{2} A} = \frac{2 s e c A}{t a n A} = \frac{2/ c o s A}{s i n A / c o s A} = \frac{2}{s i n A} = 2 cosec A$ .
Q3 · 3 marks · Short answerHarder board level
Prove that $(sin A + cosec A)^{2} + (cos A + sec A)^{2} = 7 + tan^{2} A + cot^{2} A$ , where $A$ is acute.
Hint
Expand both squares. Note that $sin A \cdot cosec A = 1$ and $cos A \cdot sec A = 1$ .
Answer
LHS $= sin^{2} A + 2 + cosec^{2} A + cos^{2} A + 2 + sec^{2} A = 1 + 4 + (1 + cot^{2} A) + (1 + tan^{2} A) = 7 + tan^{2} A + cot^{2} A$ .
Q4 · 3 marks · Short answerHarder board level
If $cosec θ - cot θ = \frac{1}{3}$ , where $θ$ is acute, find $sin θ$ and $cos θ$ . Hence find the value of $\frac{s e c θ + t a n θ}{s e c θ - t a n θ}$ .
Hint
Use $(cosec θ - cot θ) (cosec θ + cot θ) = 1$ to get $cosec θ + cot θ$ first.
Answer
From the identity, $cosec θ + cot θ = 3$ . Adding and subtracting the two equations gives $cosec θ = \frac{5}{3}$ and $cot θ = \frac{4}{3}$ . So $sin θ = \frac{3}{5}$ and $cos θ = cot θ sin θ = \frac{4}{5}$ . Then $sec θ = \frac{5}{4}$ and $tan θ = \frac{3}{4}$ , so the value is $\frac{5/4 + 3/4}{5/4 - 3/4} = \frac{2}{1/2} = 4$ .
Q5 · 3 marks · Short answerHOTS / competency
If $sin θ + sin^{2} θ = 1$ for an acute angle $θ$ , prove that $cos^{2} θ + cos^{4} θ = 1$ . Explain the key idea in one line.
Hint
Rearrange the given condition to find what $sin θ$ equals.
Answer
From the condition, $sin θ = 1 - sin^{2} θ = cos^{2} θ$ . Then $cos^{4} θ = sin^{2} θ$ , so $cos^{2} θ + cos^{4} θ = sin θ + sin^{2} θ = 1$ . Key idea: the condition says $sin θ$ equals $cos^{2} θ$ , so you can swap one for the other.
Q6 · 1 mark · MCQBoard level
MCQ: For an acute angle $θ$ , $(1 + tan^{2} θ) (1 - sin θ) (1 + sin θ)$ equals
(a) $1$ (b) $0$ (c) $sec^{2} θ$ (d) $tan^{2} θ$
Hint
Replace each bracket using an identity before multiplying.
Answer
(a) $1$ . $(1 + tan^{2} θ) = sec^{2} θ$ and $(1 - sin θ) (1 + sin θ) = 1 - sin^{2} θ = cos^{2} θ$ , so the product is $sec^{2} θ cos^{2} θ = 1$ .
Q7 · 1 mark · Assertion-reasonHarder board level
Assertion (A): There is an acute angle $θ$ for which $sec θ + tan θ = \frac{1}{2}$ .
Reason (R): $(sec θ + tan θ) (sec θ - tan θ) = 1$ wherever both ratios are defined.
(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.
Hint
Use R to find $sec θ - tan θ$ , then solve for $tan θ$ and check its sign.
Answer
(d). R is true, since the product is $sec^{2} θ - tan^{2} θ = 1$ . If A held, then $sec θ - tan θ = 2$ , so $2 tan θ = \frac{1}{2} - 2$ and $tan θ = - \frac{3}{4}$ . But $tan θ > 0$ for an acute angle (and $sec θ \geq 1$ means $sec θ + tan θ \geq 1$ anyway), so A is false.
Q8 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a straight slide in a park makes an acute angle $θ$ with the level ground, and its builder records that $cosec θ + cot θ = 3$ . The slide is 4 m long.
(i) Find $cosec θ - cot θ$ . (1 mark)
(ii) Find $sin θ$ and $cos θ$ . (1 mark)
(iii) Find the height of the top of the slide above the ground and the horizontal distance it covers, and check your answers with Pythagoras. (2 marks)
Hint
The cosec and cot partners multiply to 1. Then add and subtract the two equations.
Answer
(i) $(cosec θ + cot θ) (cosec θ - cot θ) = cosec^{2} θ - cot^{2} θ = 1$ , so $cosec θ - cot θ = \frac{1}{3}$ . (1)
(ii) Adding: $2 cosec θ = \frac{10}{3}$ , so $cosec θ = \frac{5}{3}$ and $sin θ = \frac{3}{5}$ . Subtracting: $2 cot θ = \frac{8}{3}$ , so $cot θ = \frac{4}{3}$ and $cos θ = cot θ sin θ = \frac{4}{5}$ . (1)
(iii) Height $= 4 sin θ = 4 \times \frac{3}{5} = 2.4$ m. Horizontal distance $= 4 cos θ = 4 \times \frac{4}{5} = 3.2$ m. Check: $2. 4^{2} + 3. 2^{2} = 5.76 + 10.24 = 16 = 4^{2}$ . (2)

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.What does $1 - cos^{2} A$ equal?Show answer
$sin^{2} A$
2.What does $sec^{2} A - tan^{2} A$ equal?Show answer
$1$
3.If $sec A + tan A = 4$ , what is $sec A - tan A$ ?Show answer
$\frac{1}{4}$
4.What is $cosec^{2} A - 1$ in simplest form?Show answer
$cot^{2} A$
5.What is $(1 - sin A) (1 + sin A)$ ?Show answer
$cos^{2} A$
6.Can you cross-multiply when proving an identity?Show answer
No. Transform one side until it becomes the other side.

Proving Simple Trigonometric Identities: frequently asked questions

Which side should I start from when proving an identity?

Start from the side that looks more complicated, usually the one with fractions, squares or more terms. Simplifying is easier than building up. If both sides look complicated, simplify each one separately until they meet.

Is it acceptable to work on both sides at once?

Yes, as long as you simplify each side separately and show that both reach the same expression. What you must not do is move terms across the equals sign or cross-multiply, because that assumes the result.

Do I need ratios of complementary angles for these proofs?

No. That topic is not in the 2026-27 syllabus. Use only the relations and identities on this page.

How hard are the identity questions in the board exam?

The syllabus says only simple identities are to be given. Expect proofs that use the reciprocal and quotient relations, sin squared plus cos squared equals one and its two partners, along with basic algebra such as LCM and factorising.

Do I need to write conditions like A is acute?

It is good practice. Mention the condition whenever you cancel a factor that could be zero, or whenever you take a square root. It shows that the step is valid and protects your marks.