MathsCBSE Class 10 · Introduction to Trigonometry

Identity sin²A + cos²A = 1: Proof and Applications

Q: How do I prove sin squared A plus cos squared A equals 1?

Take a triangle right-angled at B and write Pythagoras: AB squared plus BC squared equals AC squared. Divide every term by AC squared. The two fractions become cos A squared and sin A squared, giving the identity.

Q: Why is it called an identity and not an equation?

An equation is true only for particular angles, but an identity is true for every angle where both sides are defined. This one holds for all angles from 0 to 90 degrees.

Q: Why does 1 plus tan squared A equal sec squared A fail at 90 degrees?

At 90 degrees cos A is 0, and both tan A and sec A have cos A in the denominator. Division by zero is not defined, so the identity cannot be used there.

Q: Should I take a negative square root when finding cos A?

Not in Class 10. All angles here are between 0 and 90 degrees, where every ratio that exists is positive or zero, so you take the positive root.

Q: Can I use $\sin^2 A + \cos^2 A = 1$ for $A = 0^\circ$ and $A = 90^\circ$?

Yes. It holds for every angle from 0^\circ to 90^\circ: at 0^\circ it reads 0 + 1 = 1 and at 90^\circ it reads 1 + 0 = 1. The partner identities are different: 1 + \tan^2 A = \sec^2 A fails at 90^\circ and \text{cosec}^2 A = 1 + \cot^2 A fails at 0^\circ, because a ratio in them is not defined there.

Updated 14 October 2026 · Fact-checked

The identity sin squared A plus cos squared A equals 1 comes from dividing the Pythagoras theorem by the square of the hypotenuse. It holds for every angle from 0 to 90 degrees. The board asks you to prove it, derive its partners with tan, sec, cot and cosec, find one ratio from another, and use it in short proofs.

Understand Identity sin²A + cos²A = 1

Start with a right-angled triangle $A B C$ with $∠ B = 9 0^{\circ}$ . Pythagoras theorem says $A B^{2} + B C^{2} = A C^{2}$ . This is a fact about lengths. The identity turns it into a fact about angles.

Divide every term by $A C^{2}$ (allowed because $A C \neq = 0$ ): $(\frac{A B}{A C})^{2} + (\frac{B C}{A C})^{2} = 1$ . Relative to $∠ A$ , $\frac{A B}{A C} = cos A$ and $\frac{B C}{A C} = sin A$ . So $cos^{2} A + sin^{2} A = 1$ . Here $sin^{2} A$ means $(sin A)^{2}$ , not $sin (A^{2})$ .

Why call it an identity? An equation like $2 sin A = 1$ is true only for some angles. An identity is true for every angle where both sides are defined. The proof works for any acute angle, and you can check the ends: at $0^{\circ}$ , $0^{2} + 1^{2} = 1$ ; at $9 0^{\circ}$ , $1^{2} + 0^{2} = 1$ . So it holds for $0^{\circ} \leq A \leq 9 0^{\circ}$ .

Divide the same Pythagoras equation by $A B^{2}$ instead and you get $1 + tan^{2} A = sec^{2} A$ . This needs $A B \neq = 0$ , that is $cos A \neq = 0$ , so it holds for $0^{\circ} \leq A < 9 0^{\circ}$ . Divide by $B C^{2}$ and you get $cot^{2} A + 1 = cosec^{2} A$ , which needs $sin A \neq = 0$ , so it holds for $0^{\circ} < A \leq 9 0^{\circ}$ . You can also get both partners by dividing $sin^{2} A + cos^{2} A = 1$ by $cos^{2} A$ or $sin^{2} A$ .

The main use is finding one ratio from another without drawing a triangle. If $sin A = \frac{3}{5}$ and $A$ is acute, then $cos^{2} A = 1 - \frac{9}{25} = \frac{16}{25}$ , so $cos A = \frac{4}{5}$ . Take the positive root because ratios of an acute angle are positive. Similarly, from $sec A$ you get $tan A$ through $tan^{2} A = sec^{2} A - 1$ .

The identity also powers the next topic, proving simple identities, where you replace $1 - sin^{2} A$ by $cos^{2} A$ , or $sec^{2} A - tan^{2} A$ by $1$ . The factorised form $(sec A - tan A) (sec A + tan A) = 1$ is especially useful when a sum like $sec A + tan A$ is given.

Identity sin²A + cos²A = 1: formulas to remember

Fundamental identity: $sin^{2} A + cos^{2} A = 1, 0^{\circ} \leq A \leq 9 0^{\circ}$; From Pythagoras divided by the square of the hypotenuse.
Rearranged forms: $sin^{2} A = 1 - cos^{2} A, cos^{2} A = 1 - sin^{2} A$; For an acute angle take the positive square root: cos A = √(1 − sin²A).
Identity with sec and tan: $1 + tan^{2} A = sec^{2} A, 0^{\circ} \leq A < 9 0^{\circ}$; Divide by cos²A; fails at 90° where tan and sec are not defined.
Identity with cosec and cot: $1 + cot^{2} A = cosec^{2} A, 0^{\circ} < A \leq 9 0^{\circ}$; Divide by sin²A; fails at 0° where cot and cosec are not defined.
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.
Product of sin and cos from a sum: $(sin A + cos A)^{2} = 1 + 2 sin A cos A$; Square a given sum, then use sin²A + cos²A = 1.

How to solve Identity sin²A + cos²A = 1 questions

Almost every board question on this identity is either a proof of the identity itself, finding a ratio from another, or a short evaluation or condition question. This method covers all of them.

1Read what is given and what is asked. Note whether the angle is acute, since that decides the sign of square roots.
2If asked to prove the identity, draw $△ A B C$ with $∠ B = 9 0^{\circ}$ , write Pythagoras, and divide by $A C^{2}$ (or $A B^{2}$ , $B C^{2}$ for the partners), naming each ratio.
3If one ratio is given, pick the identity that links it to the one you need: sin with cos, sec with tan, cosec with cot.
4Substitute, simplify the fraction fully, then take the positive square root for an acute angle.
5Use reciprocal and quotient relations to get any remaining ratio.
6If a sum such as $sec A + tan A$ or $sin A + cos A$ is given, use the factorised form or square the sum.
7Check your answer: substitute your values back into $sin^{2} A + cos^{2} A = 1$ , and confirm sin and cos lie between 0 and 1.

Identity sin²A + cos²A = 1 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.

The identity sin²A + cos²A = 1 and its two partners

Imagine a straight footpath up a hill near your town. It is 125 m long and rises at an angle $A$ to the horizontal, where $sin A = \frac{7}{25}$ . A second straight path up the same hill rises at an angle $B$ with $cos B = \frac{4}{5}$ .

Using only the identities, find $cos A$ and $tan A$ , the height gained along the first path and the horizontal distance it covers. Then find $sin B$ and $tan B$ , and decide which path is steeper.

Try it first, then show the solutionSolution

$cos^{2} A = 1 - sin^{2} A = 1 - \frac{49}{625} = \frac{576}{625}$ ; $A$ is acute, so $cos A = \frac{24}{25}$ .
$tan A = \frac{s i n A}{c o s A} = \frac{7/25}{24/25} = \frac{7}{24}$ .
Height gained $= 125 \times \frac{7}{25} = 35$ m; horizontal distance $= 125 \times \frac{24}{25} = 120$ m. Check: $3 5^{2} + 12 0^{2} = 1225 + 14400 = 15625 = 12 5^{2}$ .
For the second path, $sin^{2} B = 1 - \frac{16}{25} = \frac{9}{25}$ , so $sin B = \frac{3}{5}$ and $tan B = \frac{3/5}{4/5} = \frac{3}{4}$ .
Steepness is the rise per metre of horizontal distance, that is $tan$ of the angle. $tan B = \frac{3}{4} = 0.75$ is greater than $tan A = \frac{7}{24} \approx 0.29$ , so the second path is steeper.

Answer: $cos A = \frac{24}{25}$ , $tan A = \frac{7}{24}$ ; height 35 m, horizontal distance 120 m. $sin B = \frac{3}{5}$ , $tan B = \frac{3}{4}$ , so the second path is steeper.

Memory hook: Know one ratio, the identity gives the other; compare slopes with tan.

The identity sin²A + cos²A = 1 and its two partners

Suppose a kite string is tied to a peg on a flat school ground. A teacher says the string makes an angle $A$ with the ground with $sec A + tan A = 3$ , and the string, pulled straight, is 50 m long.

Find $sec A$ , $tan A$ and $cos A$ , and hence the height of the kite above the ground and its horizontal distance from the peg.

Try it first, then show the solutionSolution

$(sec A - tan A) (sec A + tan A) = 1$ , so $sec A - tan A = \frac{1}{3}$ .
Adding: $2 sec A = 3 + \frac{1}{3} = \frac{10}{3}$ , so $sec A = \frac{5}{3}$ .
Subtracting: $2 tan A = 3 - \frac{1}{3} = \frac{8}{3}$ , so $tan A = \frac{4}{3}$ .
$cos A = \frac{3}{5}$ and $sin A = tan A cos A = \frac{4}{5}$ .
Height $= 50 \times \frac{4}{5} = 40$ m; horizontal distance $= 50 \times \frac{3}{5} = 30$ m.
Check: $sec^{2} A - tan^{2} A = \frac{25}{9} - \frac{16}{9} = 1$ .

Answer: $sec A = \frac{5}{3}$ , $tan A = \frac{4}{3}$ , $cos A = \frac{3}{5}$ ; height 40 m, horizontal distance 30 m.

Memory hook: Kite trick: sec plus tan times sec minus tan is 1, so one sum gives both.

Tricks to remember Identity sin²A + cos²A = 1

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

Divide by the side you want to vanish
Pythagoras $A B^{2} + B C^{2} = A C^{2}$ : divide by $A C^{2}$ for sin-cos, by $A B^{2}$ for tan-sec, by $B C^{2}$ for cot-cosec.
Use it when: Rebuild all three identities in seconds during a proof.
Sec and cosec are the big ones
In $sec^{2} A - tan^{2} A = 1$ and $cosec^{2} A - cot^{2} A = 1$ , the sec and cosec squares come first, because they are never smaller than 1. In $1 + tan^{2} A = sec^{2} A$ , $sec^{2} A$ is the big one on its own side.
Use it when: Avoid writing $tan^{2} A - sec^{2} A = 1$ by mistake.
Domain by denominator
Where you divide by $cos$ , drop $9 0^{\circ}$ ; where you divide by $sin$ , drop $0^{\circ}$ .
Use it when: State the correct range for each partner identity.

Identity sin²A + cos²A = 1: solved examples

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

Example 1: Find cos and cot from sin · 2 marks · Very short answerBoard level

If $sin A = \frac{8}{17}$ , where $A$ is acute, use the identity $sin^{2} A + cos^{2} A = 1$ to find $cos A$ and $cot A$ , and hence evaluate $cos^{2} A - sin^{2} A$ .

Try it first, then show the solutionSolution

$cos^{2} A = 1 - \frac{64}{289} = \frac{225}{289}$ , so $cos A = \frac{15}{17}$ (positive, $A$ acute). (½ mark)
$cot A = \frac{c o s A}{s i n A} = \frac{15}{8}$ . (½ mark)
$cos^{2} A - sin^{2} A = \frac{225}{289} - \frac{64}{289} = \frac{161}{289}$ . (1 mark)

Answer: $cos A = \frac{15}{17}$ , $cot A = \frac{15}{8}$ , $cos^{2} A - sin^{2} A = \frac{161}{289}$ .

Example 2: Prove the identity and a partner · 3 marks · Short answerBoard level

In $△ P QR$ , right-angled at $Q$ , prove that $sin^{2} P + cos^{2} P = 1$ . Hence prove that $1 + cot^{2} P = cosec^{2} P$ and state the angles for which it is true.

Try it first, then show the solutionSolution

By Pythagoras, $P Q^{2} + Q R^{2} = P R^{2}$ . (½ mark)
Divide by $P R^{2}$ : $(\frac{QR}{P R})^{2} + (\frac{P Q}{P R})^{2} = 1$ . (½ mark)
Relative to $∠ P$ , $\frac{QR}{P R} = sin P$ and $\frac{P Q}{P R} = cos P$ , so $sin^{2} P + cos^{2} P = 1$ . (½ mark)
Divide by $sin^{2} P$ (needs $sin P \neq = 0$ ): $1 + \frac{c o s ^{2} P}{s i n ^{2} P} = \frac{1}{s i n ^{2} P}$ . (½ mark)
So $1 + cot^{2} P = cosec^{2} P$ . (½ mark)
$sin P = 0$ only at $0^{\circ}$ , so it holds for $0^{\circ} < P \leq 9 0^{\circ}$ . (½ mark)

Answer: Both identities proved; $1 + cot^{2} P = cosec^{2} P$ holds for $0^{\circ} < P \leq 9 0^{\circ}$ .

Example 3: Express all ratios in terms of sec · 3 marks · Short answerBoard level

If $sec θ = \frac{25}{7}$ with $θ$ acute, use identities (not a triangle) to find $tan θ$ , $sin θ$ and then evaluate $\frac{1 - t a n ^{2} θ}{1 + t a n ^{2} θ}$ .

Try it first, then show the solutionSolution

$tan^{2} θ = sec^{2} θ - 1 = \frac{625}{49} - 1 = \frac{576}{49}$ , so $tan θ = \frac{24}{7}$ . (1 mark)
$cos θ = \frac{7}{25}$ , so $sin θ = tan θ cos θ = \frac{24}{25}$ . (½ mark)
Check: $\frac{576}{625} + \frac{49}{625} = 1$ . (½ mark)
$\frac{1 - \frac{576}{49}}{1 + \frac{576}{49}} = \frac{49 - 576}{49 + 576} = - \frac{527}{625}$ . (1 mark)

Answer: $tan θ = \frac{24}{7}$ , $sin θ = \frac{24}{25}$ , value $= - \frac{527}{625}$ .

Example 4: Condition sin + cos given · 5 marks · Long answerHarder board level

For an acute angle $A$ , $sin A + cos A = \frac{7}{5}$ .
(i) Find $sin A cos A$ .
(ii) Find $(sin A - cos A)^{2}$ and hence $sin A$ and $cos A$ , given $sin A > cos A$ .
(iii) Find $tan A + cot A$ in two ways and check they agree.

Try it first, then show the solutionSolution

Square: $sin^{2} A + cos^{2} A + 2 sin A cos A = \frac{49}{25}$ . (½ mark)
Use the identity: $1 + 2 sin A cos A = \frac{49}{25}$ , so $sin A cos A = \frac{12}{25}$ . (1 mark)
$(sin A - cos A)^{2} = 1 - 2 \cdot \frac{12}{25} = \frac{1}{25}$ . (½ mark)
$sin A > cos A$ , so $sin A - cos A = \frac{1}{5}$ . (½ mark)
Add and subtract with $sin A + cos A = \frac{7}{5}$ : $sin A = \frac{4}{5}$ , $cos A = \frac{3}{5}$ . (1 mark)
Way 1: $tan A + cot A = \frac{4}{3} + \frac{3}{4} = \frac{25}{12}$ . (½ mark)
Way 2: $tan A + cot A = \frac{s i n ^{2} A + c o s ^{2} A}{s i n A c o s A} = \frac{1}{12/25} = \frac{25}{12}$ ; both agree. (1 mark)

Answer: (i) $\frac{12}{25}$ (ii) $\frac{1}{25}$ ; $sin A = \frac{4}{5}$ , $cos A = \frac{3}{5}$ (iii) $\frac{25}{12}$ .

Example 5: Find the error · 4 marks · Case-basedHarder board level

Riya is told $cos A = \frac{2}{3}$ for an acute angle $A$ . She writes: " $sin A = 1 - cos A = \frac{1}{3}$ , so $tan A = \frac{1}{2}$ ."
(i) Show her values do not satisfy $sin^{2} A + cos^{2} A = 1$ .
(ii) Find the correct $sin A$ and $tan A$ .
(iii) Verify $1 + tan^{2} A = sec^{2} A$ with your values.

Try it first, then show the solutionSolution

(i) $(\frac{1}{3})^{2} + (\frac{2}{3})^{2} = \frac{5}{9} \neq = 1$ ; she used $sin A + cos A = 1$ , which is false. (1 mark)
(ii) $sin^{2} A = 1 - \frac{4}{9} = \frac{5}{9}$ , so $sin A = \frac{5}{3}$ . (½ mark)
$tan A = \frac{5 /3}{2/3} = \frac{5}{2}$ . (½ mark)
(iii) LHS $= 1 + \frac{5}{4} = \frac{9}{4}$ . (1 mark)
RHS $= (\frac{3}{2})^{2} = \frac{9}{4}$ ; equal. (1 mark)

Answer: Her error: squares dropped. Correct $sin A = \frac{5}{3}$ , $tan A = \frac{5}{2}$ ; both sides equal $\frac{9}{4}$ .

Example 6: Using (sec A − tan A)(sec A + tan A) = 1 · 3 marks · Short answerHarder board level

If $sec A + tan A = \frac{5}{2}$ for an acute angle $A$ , find $sec A - tan A$ , and hence find $sin A$ and $cos A$ .

Try it first, then show the solutionSolution

From $sec^{2} A - tan^{2} A = 1$ : $(sec A - tan A) (sec A + tan A) = 1$ , so $sec A - tan A = \frac{2}{5}$ . (1 mark)
Adding: $2 sec A = \frac{5}{2} + \frac{2}{5} = \frac{29}{10}$ , so $sec A = \frac{29}{20}$ and $cos A = \frac{20}{29}$ . (½ mark)
Subtracting: $2 tan A = \frac{5}{2} - \frac{2}{5} = \frac{21}{10}$ , so $tan A = \frac{21}{20}$ . (½ mark)
$sin A = tan A \times cos A = \frac{21}{20} \times \frac{20}{29} = \frac{21}{29}$ . Check: $\frac{441 + 400}{841} = 1$ . (1 mark)

Answer: $sec A - tan A = \frac{2}{5}$ ; $sin A = \frac{21}{29}$ , $cos A = \frac{20}{29}$ .

Example 7: MCQ: combining two identities · 1 mark · MCQBoard level

If $sin A = \frac{3}{5}$ for an acute angle $A$ , then the value of $(sec^{2} A - tan^{2} A) + cos^{2} A$ is
(a) $\frac{41}{25}$
(b) $1$
(c) $\frac{9}{25}$
(d) $\frac{34}{25}$

Try it first, then show the solutionSolution

$sec^{2} A - tan^{2} A = 1$ for every acute angle, and $cos^{2} A = 1 - \frac{9}{25} = \frac{16}{25}$ , so the value is $1 + \frac{16}{25} = \frac{41}{25}$ . (1 mark)

Answer: (a) $\frac{41}{25}$

Example 8: Assertion-reason: where an identity holds · 1 mark · Assertion-reasonBoard level

Assertion (A): $1 + tan^{2} A = sec^{2} A$ is true for $A = 9 0^{\circ}$ .
Reason (R): $tan 9 0^{\circ}$ is not 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.

Try it first, then show the solutionSolution

At $A = 9 0^{\circ}$ , $cos A = 0$ , so $tan A = \frac{s i n A}{c o s A}$ and $sec A = \frac{1}{c o s A}$ are not defined. The identity holds only for $0^{\circ} \leq A < 9 0^{\circ}$ , so A is false; R is true. (1 mark)

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

Where students lose marks in Identity sin²A + cos²A = 1

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

Mistake: Writing sin A + cos A = 1
Students take square roots of each term in the identity.
Wrong working
$cos A = \frac{3}{5}$ , so $sin A = 1 - \frac{3}{5} = \frac{2}{5}$ .
Right working
$sin^{2} A = 1 - \frac{9}{25} = \frac{16}{25}$ , so $sin A = \frac{4}{5}$ .
Do this: Always work with squares, then take the root at the end.
Usually costs: Full marks for that part, and every later step.
Mistake: Squaring only the numerator
Rushing with fractions.
Wrong working
$sin A = \frac{5}{13}$ , $sin^{2} A = \frac{25}{13}$ .
Right working
$sin^{2} A = \frac{25}{169}$ , so $cos^{2} A = \frac{144}{169}$ .
Do this: Square numerator and denominator together.
Usually costs: 1 to 2 marks in a 2- or 3-mark question.
Mistake: Dividing by the wrong side in the proof
Labels are not fixed relative to the angle.
Wrong working
Dividing $A B^{2} + B C^{2} = A C^{2}$ by $A C^{2}$ and calling $\frac{A B}{A C} = sin A$ .
Right working
$\frac{A B}{A C} = cos A$ (adjacent over hypotenuse) and $\frac{B C}{A C} = sin A$ .
Do this: Mark the angle and label opposite, adjacent, hypotenuse before dividing.
Usually costs: ½ to 1 mark in a proof.
Mistake: Mixing up the partner identity
Memorising signs without the derivation.
Wrong working
$tan^{2} A = 1 + sec^{2} A$ .
Right working
$tan^{2} A = sec^{2} A - 1$ , since $sec A \geq 1$ is the larger one.
Do this: Rederive by dividing $sin^{2} A + cos^{2} A = 1$ by $cos^{2} A$ .
Usually costs: Most marks of the question.

Exam tips

In a proof of the identity, draw and label the triangle, write Pythagoras, and name each ratio as you substitute; examiners give step marks for each.
When a ratio is given for an acute angle, write "A is acute, so we take the positive root"; it shows reasoning.
State the range of angles for the partner identities if asked, with the reason (division by cos or sin).
If a question gives a sum like sec plus tan, reach for the factorised form first.
Check your final ratios by substituting into sin² plus cos² equals 1; it takes seconds.

Practice questions on Identity sin²A + cos²A = 1

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
If $cos θ = \frac{12}{13}$ with $θ$ acute, find the value of $sin^{2} θ - cos^{2} θ + tan θ$ .
Hint
Find $sin θ$ from $1 - cos^{2} θ$ first.
Answer
$sin θ = \frac{5}{13}$ , $tan θ = \frac{5}{12}$ . Value $= \frac{25 - 144}{169} + \frac{5}{12} = - \frac{119}{169} + \frac{5}{12} = \frac{- 1428 + 845}{2028} = - \frac{583}{2028}$ .
Q2 · 3 marks · Short answerHarder board level
Prove that $1 + tan^{2} A = sec^{2} A$ starting from a right triangle $A B C$ with $∠ B = 9 0^{\circ}$ , and explain why it fails at $A = 9 0^{\circ}$ . Hence find $cos A$ if $tan A = \frac{20}{21}$ , $A$ acute.
Hint
Divide Pythagoras by $A B^{2}$ , the side adjacent to $A$ .
Answer
$\frac{A B ^{2}}{A B ^{2}} + \frac{B C ^{2}}{A B ^{2}} = \frac{A C ^{2}}{A B ^{2}}$ gives $1 + tan^{2} A = sec^{2} A$ . At $9 0^{\circ}$ , $cos A = 0$ so tan and sec are not defined. $sec^{2} A = 1 + \frac{400}{441} = \frac{841}{441}$ , $sec A = \frac{29}{21}$ , $cos A = \frac{21}{29}$ .
Q3 · 3 marks · Short answerHarder board level
If $cosec θ + cot θ = 5$ , find $cosec θ - cot θ$ , and hence find $sin θ$ and $cos θ$ .
Hint
Use $(cosec θ - cot θ) (cosec θ + cot θ) = 1$ .
Answer
$cosec θ - cot θ = \frac{1}{5}$ . Adding: $cosec θ = \frac{13}{5}$ , so $sin θ = \frac{5}{13}$ . Subtracting: $cot θ = \frac{12}{5}$ , so $cos θ = cot θ sin θ = \frac{12}{13}$ . Check: $\frac{25 + 144}{169} = 1$ .
Q4 · 5 marks · Long answerHarder board level
If $3 sin θ - cos θ = 0$ for an acute angle $θ$ , find $tan θ$ , then use $1 + tan^{2} θ = sec^{2} θ$ to find $cos θ$ and $sin θ$ . Hence evaluate $\frac{s i n ^{2} θ + 2 c o s ^{2} θ}{1 + s i n ^{2} θ}$ .
Hint
Divide the given equation by $cos θ$ .
Answer
$tan θ = \frac{1}{3}$ . $sec^{2} θ = \frac{10}{9}$ , so $cos^{2} θ = \frac{9}{10}$ , $cos θ = \frac{3}{10}$ , $sin θ = \frac{1}{10}$ . Expression $= \frac{\frac{1}{10} + \frac{18}{10}}{\frac{11}{10}} = \frac{19}{11}$ .
Q5 · 4 marks · Case-basedHOTS / competency
A student claims: "For an acute angle, $sin A = 0.6$ and $cos A = 0.7$ ." Without finding the angle, decide whether this is possible. If $sin A = 0.6$ is correct, find the true $cos A$ and $sec^{2} A - tan^{2} A$ , and explain the last result.
Hint
Test the pair in $sin^{2} A + cos^{2} A$ .
Answer
$0.36 + 0.49 = 0.85 \neq = 1$ , so impossible. True $cos A = 1 - 0.36 = 0.8$ . $sec^{2} A - tan^{2} A = \frac{1}{0.64} - \frac{0.36}{0.64} = 1$ , as it must be from $1 + tan^{2} A = sec^{2} A$ for any allowed angle.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: A ladder 13 m long leans against a vertical wall on level ground, making an acute angle $A$ with the ground, where $sin A + cos A = \frac{17}{13}$ and $sin A > cos A$ .
(i) Find $sin A cos A$ . (1 mark)
(ii) Find $sin A$ and $cos A$ , and hence how high up the wall the ladder reaches and how far its foot is from the wall. (2 marks)
(iii) The foot is then pulled out until the ladder makes an angle $B$ with the ground where $tan B = \frac{3}{4}$ . By how much does the top of the ladder slide down? (1 mark)
Hint
Square the given sum and use $sin^{2} A + cos^{2} A = 1$ ; then find $(sin A - cos A)^{2}$ .
Answer
(i) $(sin A + cos A)^{2} = \frac{289}{169}$ , so $1 + 2 sin A cos A = \frac{289}{169}$ and $sin A cos A = \frac{60}{169}$ . (1)
(ii) $(sin A - cos A)^{2} = 1 - \frac{120}{169} = \frac{49}{169}$ , so $sin A - cos A = \frac{7}{13}$ (positive, as $sin A > cos A$ ). Hence $sin A = \frac{12}{13}$ and $cos A = \frac{5}{13}$ . The ladder reaches $13 \times \frac{12}{13} = 12$ m up the wall and its foot is $13 \times \frac{5}{13} = 5$ m from the wall. (2)
(iii) $sec^{2} B = 1 + \frac{9}{16} = \frac{25}{16}$ , so $cos B = \frac{4}{5}$ and $sin B = \frac{3}{5}$ . New height $= 13 \times \frac{3}{5} = 7.8$ m, so the top slides down $12 - 7.8 = 4.2$ m. (1)

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.State the fundamental identity.Show answer
$sin^{2} A + cos^{2} A = 1$ for $0^{\circ} \leq A \leq 9 0^{\circ}$ .
2.Divide it by $cos^{2} A$ : what do you get?Show answer
$tan^{2} A + 1 = sec^{2} A$ .
3.Divide it by $sin^{2} A$ : what do you get?Show answer
$1 + cot^{2} A = cosec^{2} A$ .
4.If $sec A + tan A = 4$ , what is $sec A - tan A$ ?Show answer
$\frac{1}{4}$ .
5.Range for $1 + tan^{2} A = sec^{2} A$ ?Show answer
$0^{\circ} \leq A < 9 0^{\circ}$ .
6.Value of $cosec^{2} A - cot^{2} A$ ?Show answer
$1$ .

Identity sin²A + cos²A = 1: frequently asked questions

How do I prove sin squared A plus cos squared A equals 1?

Take a triangle right-angled at B and write Pythagoras: AB squared plus BC squared equals AC squared. Divide every term by AC squared. The two fractions become cos A squared and sin A squared, giving the identity.

Why is it called an identity and not an equation?

An equation is true only for particular angles, but an identity is true for every angle where both sides are defined. This one holds for all angles from 0 to 90 degrees.

Why does 1 plus tan squared A equal sec squared A fail at 90 degrees?

At 90 degrees cos A is 0, and both tan A and sec A have cos A in the denominator. Division by zero is not defined, so the identity cannot be used there.

Should I take a negative square root when finding cos A?

Not in Class 10. All angles here are between 0 and 90 degrees, where every ratio that exists is positive or zero, so you take the positive root.

Can I use $\sin^2 A + \cos^2 A = 1$ for $A = 0^\circ$ and $A = 90^\circ$?

Yes. It holds for every angle from 0^\circ to 90^\circ: at 0^\circ it reads 0 + 1 = 1 and at 90^\circ it reads 1 + 0 = 1. The partner identities are different: 1 + \tan^2 A = \sec^2 A fails at 90^\circ and \text{cosec}^2 A = 1 + \cot^2 A fails at 0^\circ, because a ratio in them is not defined there.