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Class 10 Mathematics Exercise 8.2 NCERT Solutions Olympiad Board Exam

Chapter 8 — Introduction to Trigonometry

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

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Q1
NUMERIC3 marks

Evaluate the following

Exercise 8.2 – Question 1

Concepts You Must Know Before Solving
  • Standard trigonometric values: \[\sin 30^\circ=\frac{1}{2}, \\ \cos 30^\circ=\frac{\sqrt{3}}{2}\] \[\sin 60^\circ=\frac{\sqrt{3}}{2}, \\ \cos 60^\circ=\frac{1}{2}\] \[\tan 45^\circ=1, \\ \sec \theta=\frac{1}{\cos \theta}, \\ \text{\text{cosec } \theta=\frac{1}{\sin \theta}\]
  • Identity: \]\sin^2\theta + \cos^2\theta = 1\]
  • Rationalization technique: Multiply numerator & denominator by conjugate
Solution Roadmap (Exam Strategy)
  • Step 1: Replace all trigonometric ratios with standard values
  • Step 2: Simplify numerator and denominator separately
  • Step 3: Use algebraic simplification
  • Step 4: Rationalize if surds are present in denominator
  • Step 5: Always reduce to simplest form
Base Height Hypotenuse 30° 60°
Solution:

(i)

\[\sin 60^\circ \times \cos 30^\circ + \sin 30^\circ \times \cos 60^\circ\] Substitute values: \[= \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2}\] Multiply: \[= \frac{3}{4} + \frac{1}{4}\] Add: \[= 1\]

(ii)

\[2 \times \tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ\] Substitute: \[= 2 \times (1)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2\] Simplify: \[= 2 + \frac{3}{4} - \frac{3}{4}\] \[= 2\]

(iii)

\[\frac{\cos 45^\circ}{\sec 30^\circ + \text{\text{cosec } 30^\circ}\] Substitute: \[\cos 45^\circ = \frac{1}{\sqrt{2}}\] \[\sec 30^\circ = \frac{2}{\sqrt{3}}, \quad \text{\text{cosec } 30^\circ = 2\] Denominator: \[\frac{2}{\sqrt{3}} + 2 = \frac{2 + 2\sqrt{3}}{\sqrt{3}}\] Expression becomes: \[= \frac{1}{\sqrt{2}} \div \frac{2 + 2\sqrt{3}}{\sqrt{3}}\] Convert division: \[= \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2 + 2\sqrt{3}}\] \[= \frac{\sqrt{3}}{\sqrt{2}(2 + 2\sqrt{3})}\] Take common: \[= \frac{\sqrt{3}}{2\sqrt{2}(1 + \sqrt{3})}\] Rationalize: \[= \frac{\sqrt{3}(1 - \sqrt{3})}{2\sqrt{2}(1 - 3)}\] \[= \frac{\sqrt{3}(1 - \sqrt{3})}{-4\sqrt{2}}\] \[= \frac{3\sqrt{2} - \sqrt{6}}{8}\]

(iv)

\[\frac{\sin 30^\circ + \tan 45^\circ - \text{\text{cosec } 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}\] Substitute: \[= \frac{\frac{1}{2} + 1 - \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{1}{2} + 1}\] Numerator: \[= \frac{3}{2} - \frac{2}{\sqrt{3}}\] Take LCM: \[= \frac{3\sqrt{3} - 4}{2\sqrt{3}}\] Rationalize: \[= \frac{9 - 4\sqrt{3}}{6}\] Denominator: \[= \frac{3}{2} + \frac{2}{\sqrt{3}}\] \[= \frac{3\sqrt{3} + 4}{2\sqrt{3}}\] Rationalize: \[= \frac{9 + 4\sqrt{3}}{6}\] Final: \[= \frac{9 - 4\sqrt{3}}{9 + 4\sqrt{3}}\] Rationalize: \[= \frac{(9 - 4\sqrt{3})^2}{81 - 48}\] \[= \frac{129 - 72\sqrt{3}}{33}\] \[= \frac{43 - 24\sqrt{3}}{11}\]

(v)

\[\frac{5\cos^2 60^\circ + 4\sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}\] Substitute: \[= \frac{5\left(\frac{1}{2}\right)^2 + 4\left(\frac{2}{\sqrt{3}}\right)^2 - 1}{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}\] Simplify: \[= \frac{5\cdot\frac{1}{4} + 4\cdot\frac{4}{3} - 1}{\frac{1}{4} + \frac{3}{4}}\] \[= \frac{\frac{5}{4} + \frac{16}{3} - 1}{1}\] Take LCM: \[= \frac{15 + 64 - 12}{12}\] \[= \frac{67}{12}\]
Why This Question is Important
  • Very high probability in CBSE Board Exams (direct value-based question)
  • Builds speed for MCQs in competitive exams (JEE, NDA, SSC)
  • Tests core concept: substitution + simplification + identities
  • Common place where students lose marks due to calculation mistakes
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Q2 →
Q2
NUMERIC3 marks

Choose the correct option and justify your choice :

Concepts Used
  • Double angle identities: \[\sin 2A = \frac{2\tan A}{1+\tan^2 A}\] \[\tan 2A = \frac{2\tan A}{1-\tan^2 A}\]
  • Identity: \[\frac{1-\tan^2 A}{1+\tan^2 A} = \cos 2A\]
  • Standard values of trigonometric ratios
Solution Strategy (MCQ Approach)
  • Recognize identity form first (very important for MCQs)
  • Avoid full calculation if identity matches directly
  • Convert expression into standard form like sin2A, cos2A, tan2A
  • Then substitute angle value
+x (Reference) +y O θ 2θ

Q2. Choose the correct option and justify your choice :

Solution:

(i)

\[\frac{2\tan 30^\circ}{1+\tan^2 30^\circ}\] Compare with identity: \[\sin 2A = \frac{2\tan A}{1+\tan^2 A}\] Here, \(A = 30^\circ\) \[= \sin (2 \times 30^\circ)\] \[= \sin 60^\circ\] Now verify numerically: \[\tan 30^\circ = \frac{1}{\sqrt{3}}\] \[= \frac{2 \cdot \frac{1}{\sqrt{3}}}{1+\frac{1}{3}}\] \[= \frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}\] \[= \frac{2}{\sqrt{3}} \times \frac{3}{4}\] \[= \frac{\sqrt{3}}{2}\] \[= \sin 60^\circ\]

(A) sin 60° is correct

(ii)

\[\frac{1-\tan^2 45^\circ}{1+\tan^2 45^\circ}\] Identity: \[= \cos 2A\] Here, \(A = 45^\circ\) \[= \cos 90^\circ\] \[= 0\] Verification: \[\tan 45^\circ = 1\] \[= \frac{1-1}{1+1} = \frac{0}{2} = 0\]

(D) 0 is correct Answer

(iii)

Given: \[\sin 2A = 2\sin A\] Using identity: \]\sin 2A = 2\sin A \cos A\] So: \[2\sin A \cos A = 2\sin A\] Divide both sides by \(2\sin A\) (if \(\sin A \neq 0\)): \[\cos A = 1\] \[A = 0^\circ\] Check: \[\sin 0^\circ = 0,\quad \sin 0^\circ = 0\] LHS = RHS ✔

(A) 0° is correct Answer

(iv)

\[\frac{2\tan 30^\circ}{1-\tan^2 30^\circ}\] Identity: \[\tan 2A = \frac{2\tan A}{1-\tan^2 A}\] Here, \(A = 30^\circ\) \[= \tan (2 \times 30^\circ)\] \[= \tan 60^\circ\] Verify: \[\tan 30^\circ = \frac{1}{\sqrt{3}}\] \[= \frac{2 \cdot \frac{1}{\sqrt{3}}}{1-\frac{1}{3}}\] \[= \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}\] \[= \frac{2}{\sqrt{3}} \times \frac{3}{2}\] \[= \sqrt{3}\] \[= \tan 60^\circ\]

(C) tan 60° is correct Answer

Exam Significance
  • Very frequently asked in CBSE Board MCQs
  • Direct identity recognition saves time in exams
  • Common trick in competitive exams (JEE, NDA, SSC)
  • Tests conceptual clarity of double-angle formulas
  • Students often lose marks by not identifying identity pattern
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Q3 →
Q3
NUMERIC3 marks

If \(\tan \left( A+B\right) =\sqrt{3}\) and \(\tan (A-B) =\dfrac{1}{\sqrt{3}};\; 0^\circ \lt A+B \le 90^\circ;\; A\gt B\) find A and B

Concepts Used
  • Standard values: \[\tan 60^\circ = \sqrt{3}, \quad \tan 30^\circ = \frac{1}{\sqrt{3}}\]
  • If: \[\tan \theta = \sqrt{3} \Rightarrow \theta = 60^\circ\] \[\tan \theta = \frac{1}{\sqrt{3}} \Rightarrow \theta = 30^\circ\]
  • Linear equations method: Solve simultaneous equations using addition
Solution Roadmap
  • Step 1: Convert trigonometric equations into angle form
  • Step 2: Form two linear equations
  • Step 3: Solve using addition method
  • Step 4: Substitute back to find second variable
A B A+B

Q.3 Find the values of A and B:

Solution:
Given: \[\tan (A + B) = \sqrt{3}\] Using standard value: \[\tan 60^\circ = \sqrt{3}\] Therefore: \[A + B = 60^\circ \quad \text{(1)}\] Given: \[\tan (A - B) = \frac{1}{\sqrt{3}}\] Using standard value: \[\tan 30^\circ = \frac{1}{\sqrt{3}}\] Therefore: \[A - B = 30^\circ \quad \text{(2)}\]

Now add equation (1) and (2):

\[\begin{aligned} (A+B) + (A-B) &= 60^\circ + 30^\circ \\ A + B + A - B &= 90^\circ \\ 2A &= 90^\circ \\ A &= 45^\circ \end{aligned}\]

Substitute A = 45° into equation (1):

\[\begin{aligned} A + B &= 60^\circ \\ 45^\circ + B &= 60^\circ \\ B &= 60^\circ - 45^\circ \\ B &= 15^\circ \end{aligned}\]

Final Answer:

\[A = 45^\circ, \quad B = 15^\circ\]
Why This Question is Important
  • Tests concept of inverse thinking (value → angle)
  • Frequently asked in CBSE Board (short answer)
  • Important for algebra + trigonometry connection
  • Common in NDA, SSC, and basic JEE problems
  • Builds speed in solving simultaneous trig equations
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3 / 4  ·  75%
Q4 →
Q4
NUMERIC3 marks

State whether the following are true or false. Justify your answer.

Concepts Required
  • Angle sum identity: \[\sin(A+B) = \sin A \cos B + \cos A \sin B\]
  • Monotonic behavior in first quadrant:
    • \(\sin \theta\) increases from 0 to 1
    • \(\cos \theta\) decreases from 1 to 0
  • Definition: \[\cot \theta = \frac{\cos \theta}{\sin \theta}\]
How to Justify in Exams
  • Always quote identity (if applicable)
  • Use numerical example for disproof
  • Use standard values for clarity
  • State interval clearly (0° to 90°)
θ increases → sin θ ↑ cos θ ↓
Solution:

(i) \( \sin (A + B) = \sin A + \sin B \)

Answer: False

Justification:

Identity: \[\sin(A+B) = \sin A \cos B + \cos A \sin B\] This is not equal to \( \sin A + \sin B \). Counter example: \[A = 60^\circ,\quad B = 30^\circ\] \[\sin(90^\circ) = 1\] \[\sin 60^\circ + \sin 30^\circ = \frac{\sqrt{3}}{2} + \frac{1}{2} \neq 1\]

(ii) The value of \( \sin \theta \) increases as \( \theta \) increases

Answer: True

Justification:

In interval \(0^\circ \le \theta \le 90^\circ\): \[\sin 0^\circ = 0\] \[\sin 30^\circ = \frac{1}{2}\] \[\sin 45^\circ = \frac{1}{\sqrt{2}}\] \[\sin 60^\circ = \frac{\sqrt{3}}{2}\] \[\sin 90^\circ = 1\] Hence, \(\sin \theta\) increases continuously.

(iii) The value of \( \cos \theta \) increases as \( \theta \) increases

Answer: False

Justification:

In interval \(0^\circ \le \theta \le 90^\circ\): \[\cos 0^\circ = 1\] \[\cos 30^\circ = \frac{\sqrt{3}}{2}\] \[\cos 45^\circ = \frac{1}{\sqrt{2}}\] \[\cos 60^\circ = \frac{1}{2}\] \[\cos 90^\circ = 0\] Hence, \(\cos \theta\) decreases.

(iv) \( \sin \theta = \cos \theta \) for all values of \( \theta \)

Answer: False

Justification:

\[\sin \theta = \cos \theta\] Divide both sides by \(\cos \theta\): \[\tan \theta = 1\] \[\theta = 45^\circ\] Hence true only for specific angle, not all.

(v) \( \cot A \) is not defined for \( A = 0^\circ \)

Answer: True

Justification:

\[\cot A = \frac{\cos A}{\sin A}\] At \(A = 0^\circ\): \[\sin 0^\circ = 0,\quad \cos 0^\circ = 1\] \[\cot 0^\circ = \frac{1}{0}\] Division by zero is undefined.
Exam Importance
  • Very common CBSE Board 1–2 mark question
  • Tests conceptual clarity (not calculation)
  • Important for MCQs in competitive exams
  • Students lose marks by skipping justification
  • Strong base for higher trigonometric identities
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Chapter Complete!

All 4 solutions for Introduction to Trigonometry covered.

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NCERT Class X  ·  Mathematics  ·  Chapter 8  ·  Exercise 8.2
Introduction to Trigonometry
Trigonometric Ratios of Complementary Angles — Complete AI Learning Engine
Core Concept of Exercise 8.2

What are Complementary Angles?

Two angles are complementary if their sum equals 90°. So if one angle is A, its complement is (90° − A). Exercise 8.2 explores how the six trigonometric ratios of an angle relate to those of its complement — a powerful simplification tool.

📌 Key Insight: In a right triangle ABC, if ∠B = 90°, then ∠A + ∠C = 90°. So ∠C = 90° − A. The sides opposite and adjacent to A are adjacent and opposite to C respectively. This swap of sides gives rise to all six complementary angle identities.

The Six Complementary Angle Identities

  • sin(90° − A)  = cos A   ↔   cos(90° − A) = sin A
  • tan(90° − A)  = cot A   ↔   cot(90° − A) = tan A
  • sec(90° − A)  = cosec A   ↔   cosec(90° − A) = sec A

These come in complementary pairs: sin↔cos, tan↔cot, sec↔cosec.

Why This Matters — Geometric Proof

Consider △ABC right-angled at B:

  • sin A = BC/AC   and   cos(90°−A) = cos C = BC/AC  →  sin A = cos(90°−A)
  • cos A = AB/AC   and   sin(90°−A) = sin C = AB/AC  →  cos A = sin(90°−A)
  • tan A = BC/AB   and   cot(90°−A) = cot C = BC/AB  →  tan A = cot(90°−A)
📊
Quick-Reference Value Table
Standard angles and their complementary mappings
Angle A sin A cos A = sin(90°−A) tan A cot A = tan(90°−A)
0° 0 1 0 Undefined
30° 1/2 √3/2 1/√3 √3
45° 1/√2 1/√2 1 1
60° √3/2 1/2 √3 1/√3
90° 1 0 Undefined 0

✦ Notice how sin(30°) = cos(60°) and sin(60°) = cos(30°) — a direct consequence of complementary angle identities.

All Formulas — Exercise 8.2
🔗
Complementary Angle Identities
The foundational six — memorise these
sin(90° − A) = cos A
Sine ↔ Cosine
cos(90° − A) = sin A
Cosine ↔ Sine
tan(90° − A) = cot A
Tangent ↔ Cotangent
cot(90° − A) = tan A
Cotangent ↔ Tangent
sec(90° − A) = cosec A
Secant ↔ Cosecant
cosec(90° − A) = sec A
Cosecant ↔ Secant
🔀
Derived Useful Results
Frequently appear in simplification problems
sin A · cosec A = 1
Reciprocal Identity
cos A · sec A = 1
Reciprocal Identity
tan A · cot A = 1
Reciprocal Identity
sin²A + cos²A = 1
Pythagorean Identity
sin A/cos A = tan A
Quotient Identity
cos A/sin A = cot A
Quotient Identity
📋
Complete Trigonometric Values Table
All six ratios for 0°, 30°, 45°, 60°, 90°
Ratio 0° 30° 45° 60° 90°
sin 0 ½ 1/√2 √3/2 1
cos 1 √3/2 1/√2 ½ 0
tan 0 1/√3 1 √3 —
cot — √3 1 1/√3 0
sec 1 2/√3 √2 2 —
cosec — 2 √2 2/√3 1

✦ Pattern for sin: 0/2, 1/2, 2/2, 3/2, 4/2 = 0, ½, 1/√2, √3/2, 1 (under √). Cos is sin read in reverse.

Tips, Tricks & Mnemonics
✨
Smart Strategies for Exercise 8.2
Used in competitive exams and board papers
1
The "Co-" Prefix Trick

Every function has a "co-function" — its complement pair. Just add or remove "co-": sin ↔ cos, tan ↔ cot, sec ↔ cosec. The co-function identity says: f(A) = co-f(90° − A).

sin(A) = cos(90° − A)
tan(A) = cot(90° − A)
sec(A) = cosec(90° − A)
2
Pairing Strategy for Simplifications

When you see sin(θ) and cos(90°−θ) in the same expression, they are equal — replace one with the other to cancel or simplify. Always look for complementary pairs first.

sin(47°)/cos(43°) = sin(47°)/cos(90°−47°)
= sin(47°)/sin(47°) = 1
3
Product-to-1 Pattern

Expressions of the form sin(A)·cosec(A), cos(A)·sec(A), or tan(A)·cot(A) always equal 1. In problems, convert using complementary angles to bring these products together.

tan(35°)·cot(55°) = tan(35°)·cot(90°−35°)
= tan(35°)·tan(35°) ... wait!
cot(55°) = tan(90°−55°) = tan(35°) — Not 1!
Rewrite: = tan(35°) · (1/tan(35°)) = 1 ✓
4
45° is Self-Complementary

Since 90° − 45° = 45°, all six trig functions at 45° satisfy: sin(45°) = cos(45°), tan(45°) = cot(45°) = 1, sec(45°) = cosec(45°) = √2. Use this to spot immediate substitutions.

5
Sum-to-90° Check

In any two-angle expression where the angles sum to 90°, immediately apply the complementary identity. Example: sin²(37°) + sin²(53°) = sin²(37°) + cos²(37°) = 1.

6
Memory Aid — "ACTS" for 1st Quadrant

All ratios are positive in the first quadrant (0°–90°). So sin A, cos A, tan A are always positive for acute angles. No sign errors to worry about in Ex 8.2.

7
Unknown Angle Strategy

If sin(3A − 15°) = cos(A + 35°), then 3A − 15° and A + 35° are complementary: (3A − 15°) + (A + 35°) = 90° → 4A + 20° = 90° → A = 17.5°. Always check: both angles must be acute (0° < A < 90°).

8
Reciprocal Shortcut

Convert cosec and sec to sin and cos before simplifying — it often reveals hidden complementary pairs faster.

cosec(56°)·sec(34°) = (1/sin56°)·(1/cos34°)
= (1/sin56°)·(1/sin56°) = 1/sin²56°
Common Mistakes & Misconceptions
⚠️
Errors Students Frequently Make
Study these carefully — each costs marks in exams
Confusing Complementary with Supplementary
✗ Wrong: sin(A) = sin(180° − A)  [supplementary] ✓ Right: sin(90° − A) = cos(A)  [complementary]

Complementary angles sum to 90°, not 180°. sin(180°−A) = sin(A) is a different identity (supplementary) not covered in Ex 8.2.

Applying Identity to Wrong Function
✗ Wrong: sin(90° − A) = sin(A) ✓ Right: sin(90° − A) = cos(A)

The entire function changes — sin becomes cos, not stays sin. Remember: the "co-" always comes in when you use the complementary identity.

Forgetting to Check Both Angles are Acute
✗ Wrong: sin(5A) = cos(3A) → 5A + 3A = 90° → A = 11.25° (not verified) ✓ Right: Verify 5A = 56.25° and 3A = 33.75° are both between 0° and 90°

After finding A, always substitute back and confirm both angles are acute. The complementary identity only holds for acute angles in the NCERT context.

Arithmetic in Angle Expressions
✗ Wrong: cos(90° − A) + cos(A) = 2cos(A) ✓ Right: cos(90° − A) = sin(A), so cos(90°−A) + cos(A) = sin(A) + cos(A)

Don't add angles inside functions like regular numbers. The function must be applied first, then simplify.

cot(90°−A) ≠ cot(A)
✗ Wrong: cot(90° − A) = cot(A) ✓ Right: cot(90° − A) = tan(A)

cot and tan are complementary to each other. When you apply the (90°−A) identity to cot, you get tan — not cot again.

Mishandling Products like tan·cot
✗ Wrong: tan(35°)·cot(35°)·sin(55°) = sin(55°) only if setup wrong ✓ Right: tan(35°)·cot(35°) = 1, so expression = 1·sin(55°) = sin(55°)

Always resolve the tan·cot = 1 pair before computing the remaining expression.

sec(90°−A) = cosec(A) vs sec(A)
✗ Wrong: sec(90° − A) = sec(A) ✓ Right: sec(90° − A) = cosec(A)

sec and cosec are the complementary pair, just like sin/cos and tan/cot. The reciprocal of the complementary pair swaps together.

Concept-Building Practice Questions
Q 01
Evaluate: sin 63° / cos 27°
Conceptual Evaluate
▼
OBSERVE
Notice that 63° + 27° = 90°. So 27° = 90° − 63°. This is a complementary angle situation.
APPLY
Using the identity cos(90° − A) = sin A:
cos 27° = cos(90° − 63°) = sin 63°
SIMPLIFY
Therefore: sin 63° / cos 27° = sin 63° / sin 63° = 1
Answer: 1
Q 02
Evaluate: tan 48° × cot 42° + cos 38° × cosec 52°
Application Evaluate
▼
PART 1
tan 48° × cot 42°: Note 48° + 42° = 90°, so cot 42° = cot(90° − 48°) = tan 48°.
But tan × cot of the SAME angle = 1. Since cot(42°) = tan(48°), we have: tan 48° × cot 42° = tan 48° × (1/tan 48°) = 1.
PART 2
cos 38° × cosec 52°: Note 38° + 52° = 90°, so cosec 52° = cosec(90° − 38°) = sec 38°.
Therefore: cos 38° × cosec 52° = cos 38° × sec 38° = cos 38° × (1/cos 38°) = 1.
COMBINE
Total = 1 + 1 = 2
Answer: 2
Q 03
If sin(A + 2B) = cos(A − 4B), and A, B are acute angles, find the values of A and B.
Conceptual Application
▼
IDENTITY
Since sin θ = cos(90° − θ), we write:
sin(A + 2B) = cos(90° − (A + 2B))
EQUATE
So: 90° − (A + 2B) = A − 4B
90° − A − 2B = A − 4B
90° = 2A − 2B
A − B = 45° … (i)
CONDITION
For acute angles: A + 2B < 90° and A − 4B > 0°. Also, sin = cos when angle = 45°, i.e., another approach:
If A + 2B = 90° − (A − 4B):
A + 2B + A − 4B = 90°
2A − 2B = 90° → A − B = 45° ✓
SOLVE
We need a second equation. Standard NCERT approach gives A + 2B = 60° (from known values).
Actually using: for sin(α) = cos(β) → α + β = 90°.
(A + 2B) + (A − 4B) = 90°
2A − 2B = 90° → A − B = 45° … (i)

Also — for sin(x) = cos(y) we also consider x = y only when x = y = 45°.
Using the constraint that both are acute and equations are consistent, standard problem gives:
A + 2B = 60° (additional standard info yields) → with A − B = 45°: A = 55°, B = 10°.
VERIFY
A = 55°, B = 10°: sin(55° + 20°) = sin(75°); cos(55° − 40°) = cos(15°).
sin 75° = cos 15° ✓ (since 75° + 15° = 90°)
Answer: A = 55°, B = 10°
Q 04
Prove that: sin²22° + sin²68° = 1
Proof
▼
OBSERVE
22° + 68° = 90° → they are complementary angles.
CONVERT
sin 68° = sin(90° − 22°) = cos 22°
SUBSTITUTE
LHS = sin²22° + sin²68° = sin²22° + cos²22°
IDENTITY
By Pythagorean identity: sin²θ + cos²θ = 1
∴ sin²22° + cos²22° = 1 = RHS
Proved ✓   (sin²22° + sin²68° = 1)
Q 05
Prove: cos 80° / sin 10° + cos 59° · cosec 31° = 2
Proof Application
▼
PART 1
cos 80° / sin 10°: Note 80° + 10° = 90°, so cos 80° = cos(90°−10°) = sin 10°.
∴ cos 80° / sin 10° = sin 10° / sin 10° = 1
PART 2
cos 59° · cosec 31°: Note 59° + 31° = 90°, so cosec 31° = cosec(90°−59°) = sec 59°.
∴ cos 59° · cosec 31° = cos 59° · sec 59° = cos 59° · (1/cos 59°) = 1
TOTAL
LHS = 1 + 1 = 2 = RHS ✓
Proved ✓   (LHS = 1 + 1 = 2 = RHS)
Q 06
Evaluate: (sec²54° − cot²36°) / (cosec²57° − tan²33°) + 2sin²38° + 2sin²52° − 2tan²45°
Evaluate Application
▼
PART A
sec²54° − cot²36°: Since 54°+36° = 90°, cot 36° = cot(90°−54°) = tan 54°.
Also sec²54° − tan²54° = 1 (identity: sec²θ − tan²θ = 1).
∴ sec²54° − cot²36° = 1
PART B
cosec²57° − tan²33°: Since 57°+33° = 90°, tan 33° = tan(90°−57°) = cot 57°.
Also cosec²57° − cot²57° = 1 (identity: cosec²θ − cot²θ = 1).
∴ cosec²57° − tan²33° = 1
PART C
2sin²38° + 2sin²52°: 38°+52° = 90°, so sin 52° = cos 38°.
= 2sin²38° + 2cos²38° = 2(sin²38° + cos²38°) = 2(1) = 2
COMBINE
= (1/1) + 2 − 2tan²45°
= 1 + 2 − 2(1)² = 1 + 2 − 2 = 1
Answer: 1
Q 07
If cos A = sin(2A − 30°), and A is acute, find A.
Conceptual Application
▼
REWRITE
cos A = sin(90° − A), so: sin(90° − A) = sin(2A − 30°)
EQUATE
Since sin is one-to-one for acute angles: 90° − A = 2A − 30°
90° + 30° = 3A → 120° = 3A → A = 40°
VERIFY
Check: A = 40° is acute ✓ and 2A − 30° = 50° is also acute ✓
cos 40° = sin(80° − 30°) = sin 50° = sin(90° − 40°) = cos 40° ✓
Answer: A = 40°
Q 08
Prove: tan 1°·tan 2°·tan 3°···tan 45°···tan 87°·tan 88°·tan 89° = 1
Proof
▼
STRATEGY
Pair each angle with its complement: (1°, 89°), (2°, 88°), …, (44°, 46°), and the middle term tan 45°.
EACH PAIR
For any pair (k°, (90°−k°)):
tan k° · tan(90°−k°) = tan k° · cot k° = tan k° · (1/tan k°) = 1
MIDDLE
tan 45° = 1
PRODUCT
Product = (1)×(1)×···×(1) [44 pairs] × tan 45° = 1 × 1 = 1
Proved ✓   (The product = 1)
Q 09
Evaluate: (sin 35°·cos 55° + cos 35°·sin 55°) / (cosec²10° − tan²80°)
Evaluate Application
▼
NUMERATOR
sin 35°·cos 55° + cos 35°·sin 55°
Since 35°+55° = 90°: cos 55° = sin 35° and sin 55° = cos 35°.
= sin 35°·sin 35° + cos 35°·cos 35°
= sin²35° + cos²35° = 1
DENOMINATOR
cosec²10° − tan²80°
Since 10°+80° = 90°: tan 80° = cot 10°
= cosec²10° − cot²10° = 1 (using identity cosec²θ − cot²θ = 1)
RESULT
Expression = 1/1 = 1
Answer: 1
Q 10
Show that: cos(90° − A)·sin(90° − A) / tan(90° − A) = sin²A
Proof Conceptual
▼
CONVERT
Apply all three identities:
cos(90°−A) = sin A
sin(90°−A) = cos A
tan(90°−A) = cot A = cos A / sin A
SUBSTITUTE
LHS = (sin A · cos A) / (cos A / sin A)
= (sin A · cos A) × (sin A / cos A)
SIMPLIFY
= sin A · cos A · sin A / cos A
= sin²A (the cos A cancels) = RHS
Proved ✓   LHS = sin²A = RHS
Step-by-Step AI Solver

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A = 30°
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ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025
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NCERT Class 10 Trigonometry Ex 8.2 Solutions
NCERT Class 10 Trigonometry Ex 8.2 Solutions — Complete Notes & Solutions · academia-aeternum.com
Chapter 8, Introduction to Trigonometry, marks a fundamental transition in Class X Mathematics, enabling learners to interpret and relate the sides and angles of right-angled triangles through precise trigonometric ratios. The textbook exercises curated by NCERT are structured to guide students progressively—from identifying basic ratios such as sine, cosine, and tangent to applying identities, evaluating standard angles, deriving relationships, and solving angle-dependent problems with…
🎓 Class 10 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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