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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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Complete each identity by selecting the correct answer from the dropdown

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ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025
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Introduction To Trigonometry | Mathematics Class -10
Introduction To Trigonometry | Mathematics Class -10 — Complete Notes & Solutions · academia-aeternum.com
🎓 Class -10 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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