Exercise 8.2 — Introduction to Trigonometry

1. 16 marksNCERT Cl-10 Maths, Ex 8.2, Q1

   Evaluate: (i) sin 60° cos 30° + sin 30° cos 60°; (ii) 2 tan²45° + cos²30° − sin²60°; (iii) cos 45° / (sec 30° + cosec 30°); (iv) (sin 30° + tan 45° − cosec 60°) / (sec 30° + cos 60° + cot 45°); (v) (5 cos²60° + 4 sec²30° − tan²45°) / (sin²30° + cos²30°).

   Hint. Substitute from the table, then simplify. Parts (iii) and (iv) need rationalising at the end; part (v) has a denominator you should recognise instantly.

   (i) sin 60° cos 30° + sin 30° cos 60° = (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4

   ✦ = 1

   (ii) 2 tan²45° + cos²30° − sin²60° tan 45° = 1, so the first term is 2(1)² = 2. cos 30° = √3/2 and sin 60° = √3/2 — the same number, so cos²30° − sin²60° = 3/4 − 3/4 = 0.

   ✦ = 2

   (iii) cos 45° / (sec 30° + cosec 30°)

   Numerator: cos 45° = 1/√2 Denominator: sec 30° + cosec 30° = 2/√3 + 2 = (2 + 2√3)/√3

   So the expression is (1/√2) × √3/(2 + 2√3) = √3 / (√2 · 2(1 + √3))

   Rationalise by multiplying top and bottom by (√3 − 1), because (1 + √3)(√3 − 1) = 3 − 1 = 2: = √3(√3 − 1) / (2√2 × 2) = (3 − √3)/(4√2)

   Clear the surd in the denominator: = (3 − √3)√2 / 8 = (3√2 − √6)/8

   ✦ = (3√2 − √6)/8 ≈ 0.224

   (iv) (sin 30° + tan 45° − cosec 60°) / (sec 30° + cos 60° + cot 45°)

   Numerator: 1/2 + 1 − 2/√3 = 3/2 − 2/√3 Denominator: 2/√3 + 1/2 + 1 = 2/√3 + 3/2

   Put each over the common denominator 2√3: Numerator = (3√3 − 4)/(2√3), Denominator = (4 + 3√3)/(2√3)

   The 2√3 cancels, leaving (3√3 − 4)/(3√3 + 4). Rationalise using (3√3)² = 27: = (3√3 − 4)² / ((3√3)² − 4²) = (27 − 24√3 + 16)/(27 − 16) = (43 − 24√3)/11

   ✦ = (43 − 24√3)/11 ≈ 0.130

   (v) (5 cos²60° + 4 sec²30° − tan²45°) / (sin²30° + cos²30°)

   The denominator is sin²30° + cos²30°, which is 1 by the first identity — no need to substitute anything.

   Numerator: 5(1/2)² + 4(2/√3)² − 1² = 5/4 + 16/3 − 1 Over a common denominator of 12: 15/12 + 64/12 − 12/12 = 67/12

   ✦ = 67/12

   Where students slip. In (v), grinding out sin²30° + cos²30° as 1/4 + 3/4. It is correct but wasteful — recognising the identity is the point of putting it there.

   Another way. Part (i) is the addition formula sin(60° + 30°) = sin 90° = 1, and part (ii)'s middle terms vanish because cos 30° and sin 60° are the same value. Spotting these gives both answers with almost no work.

2. 24 marksNCERT Cl-10 Maths, Ex 8.2, Q2

   Choose the correct option and justify. (i) 2 tan 30° / (1 + tan²30°) = (A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°. (ii) (1 − tan²45°)/(1 + tan²45°) = (A) tan 90° (B) 1 (C) sin 45° (D) 0. (iii) sin 2A = 2 sin A is true when A = (A) 0° (B) 30° (C) 45° (D) 60°. (iv) 2 tan 30° / (1 − tan²30°) = (A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°.

   Hint. Substitute tan 30° = 1/√3 and simplify, then match the number against the table. Part (iii) needs factoring, not substitution.

   (i) With tan 30° = 1/√3, tan²30° = 1/3. 2(1/√3) / (1 + 1/3) = (2/√3) ÷ (4/3) = (2/√3)(3/4) = 3/(2√3) = √3/2

   And √3/2 is sin 60°.

   ✦ (A) sin 60°

   (ii) tan 45° = 1, so tan²45° = 1. (1 − 1)/(1 + 1) = 0/2 = 0

   ✦ (D) 0

   (iii) Do not test the options one by one — factor instead. sin 2A = 2 sin A cos A, so the equation is 2 sin A cos A = 2 sin A 2 sin A (cos A − 1) = 0

   So either sin A = 0 or cos A = 1. Both happen at A = 0°, and for 0° ≤ A ≤ 90° there is no other solution, since cos A = 1 only at 0°.

   Checking: at A = 0°, sin 0° = 0 and 2 sin 0° = 0 ✓

   ✦ (A) 0°

   (iv) Same substitution as (i), but the sign in the denominator is different. 2(1/√3) / (1 − 1/3) = (2/√3) ÷ (2/3) = (2/√3)(3/2) = 3/√3 = √3

   And √3 is tan 60°.

   ✦ (C) tan 60°

   Where students slip. Reading sin 2A as 2 × sin A in part (iii) and concluding the statement is always true. sin 2A means the sine of the doubled angle; the two agree only where the factored equation says they do.

   Another way. Parts (i) and (iv) are the Class 11 double-angle formulas for sin 2θ and tan 2θ with θ = 30°, which is why the answers are sin 60° and tan 60°. You are not expected to know that, but it explains the pattern.

3. 33 marksNCERT Cl-10 Maths, Ex 8.2, Q3

   If tan(A + B) = √3 and tan(A − B) = 1/√3, where 0° < A + B ≤ 90° and A > B, find A and B.

   Hint. Read each tangent value off the table to get two equations in A and B, then solve them as a simple pair.

   Step 1 — Convert each equation to an angle. tan(A + B) = √3, and tan 60° = √3, so A + B = 60° … (1)

   tan(A − B) = 1/√3, and tan 30° = 1/√3, so A − B = 30° … (2)

   The condition 0° < A + B ≤ 90° is what makes these readings unique — without it, tangent repeats and other angles would qualify.

   Step 2 — Add (1) and (2) to eliminate B. 2A = 90°, so A = 45°

   Step 3 — Substitute back into (1). 45° + B = 60°, so B = 15°

   Check. A + B = 60° ✓, A − B = 30° ✓, and A > B as required ✓

   ✦ Answer: A = 45° and B = 15°

   Where students slip. Trying to expand tan(A + B) with an addition formula. That formula is Class 11 material and is not needed — the whole method is to recognise √3 and 1/√3 as table values.

   Another way. Subtract instead of adding: (1) − (2) gives 2B = 30°, so B = 15°, and then A = 60° − 15° = 45°. Same pair of equations, either order.

4. 45 marksNCERT Cl-10 Maths, Ex 8.2, Q4

   State whether each is true or false, and justify. (i) sin(A + B) = sin A + sin B. (ii) The value of sin θ increases as θ increases. (iii) The value of cos θ increases as θ increases. (iv) sin θ = cos θ for all values of θ. (v) cot A is not defined for A = 0°.

   Hint. For the false statements, one counter-example from the table is a complete justification. Take θ to run from 0° to 90° throughout.

   (i) False. Take A = B = 30°. Then sin(A + B) = sin 60° = √3/2 ≈ 0.87, but sin A + sin B = 1/2 + 1/2 = 1. The two are different, so the statement fails. (Sine is not a linear function, so it does not distribute over a sum.)

   (ii) True. As θ grows from 0° to 90°, the table reads 0, 1/2, 1/√2, √3/2, 1 — increasing at every step. Geometrically, keeping the hypotenuse fixed and opening the angle lengthens the opposite side, so opposite/hypotenuse grows.

   (iii) False. cos θ runs 1, √3/2, 1/√2, 1/2, 0 across the same angles — it decreases. Opening the angle shortens the adjacent side while the hypotenuse stays fixed, so the ratio falls.

   (iv) False. At θ = 30°, sin 30° = 1/2 but cos 30° = √3/2, and these are not equal. The two are equal at exactly one angle in the range, θ = 45°, where both are 1/√2 — 'for all values' is what makes the statement false.

   (v) True. cot A = cos A / sin A, and sin 0° = 0. Division by zero is undefined, so cot 0° is undefined. Read from the sides, cot A = adjacent/opposite, and at 0° the opposite side has shrunk to nothing.

   ✦ Answer: (i) False (ii) True (iii) False (iv) False (v) True

   Where students slip. Answering (v) with 'cot 0° = ∞'. Infinity is not a value, and the textbook is explicit that these ratios are *not defined* — writing ∞ is marked wrong.

   Another way. Statements (ii) and (iii) are two halves of one fact: as the angle opens, the opposite side grows and the adjacent side shrinks. Holding that picture answers both without consulting the table.

Solutions written by the tuition.in editorial team and checked against the NCERT Class 10 Mathematics textbook, Reprint 2026-27 (chapter 8 now runs to three exercises — 8.1, 8.2 and 8.3; trigonometric ratios of complementary angles were removed from this chapter, and the summary's six points make no mention of them). Questions are referenced from the NCERT textbook for identification.