Exercise 8.2 answers: Introduction to Trigonometry

Open NCERT p. 127Matches NCERT’s answer

(i) sin 60° cos 30° + sin 30° cos 60°

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

Answer1

(ii) 2 tan² 45° + cos² 30° – sin² 60°

1. tan 45° = 1, so 2 tan² 45° = 2.
2. cos² 30° = (√3/2)² = 3/4 and sin² 60° = (√3/2)² = 3/4.
3. 2 + 3/4 − 3/4 = 2.

Answer2

(iii) cos 45°/(sec 30° + cosec 30°)

1. cos 45° = 1/√2, sec 30° = 2/√3 and cosec 30° = 2.
2. Bottom: 2/√3 + 2 = (2 + 2√3)/√3 = 2(1 + √3)/√3.
3. So the expression = (1/√2) × √3/(2(1 + √3)) = √3/(2√2(1 + √3)).
4. Multiply top and bottom by (√3 − 1): the bottom becomes 2√2(3 − 1) = 4√2 and the top becomes √3(√3 − 1) = 3 − √3.
5. So the value is (3 − √3)/(4√2). Multiplying top and bottom by √2 gives (3√2 − √6)/8.

Answer(3√2 − √6)/8

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

1. sin 30° = 1/2, tan 45° = 1, cosec 60° = 2/√3, sec 30° = 2/√3, cos 60° = 1/2, cot 45° = 1.
2. Top: 1/2 + 1 − 2/√3 = 3/2 − 2/√3 = (3√3 − 4)/(2√3).
3. Bottom: 2/√3 + 1/2 + 1 = 3/2 + 2/√3 = (3√3 + 4)/(2√3).
4. Dividing, the 2√3 cancels: the value is (3√3 − 4)/(3√3 + 4).
5. Multiply top and bottom by (3√3 − 4): the bottom becomes (3√3)² − 4² = 27 − 16 = 11 and the top becomes (3√3 − 4)² = 27 − 24√3 + 16 = 43 − 24√3.
6. So the value is (43 − 24√3)/11.

Answer(43 − 24√3)/11

(v) (5 cos² 60° + 4 sec² 30° – tan² 45°)/(sin² 30° + cos² 30°)

1. sin² 30° + cos² 30° = 1/4 + 3/4 = 1, so the bottom is 1.
2. cos² 60° = 1/4, so 5 cos² 60° = 5/4.
3. sec² 30° = 4/3, so 4 sec² 30° = 16/3; tan² 45° = 1.
4. 5/4 + 16/3 − 1 = 15/12 + 64/12 − 12/12 = 67/12.

Answer67/12

Watch this explained “All thirty cells”, 8:14 into The extreme cases at 0° and 90°, and the ratios that stop being defined

Open NCERT p. 127Matches NCERT’s answer

(i) 2 tan 30°/(1 + tan² 30°)

1. tan 30° = 1/√3, so tan² 30° = 1/3 and 1 + tan² 30° = 4/3.
2. 2 tan 30° = 2/√3, so the expression = (2/√3) ÷ (4/3) = (2/√3) × (3/4) = 3/(2√3) = √3/2.
3. sin 60° = √3/2, while cos 60° = 1/2, tan 60° = √3 and sin 30° = 1/2, so only (A) matches.

Answer(A) sin 60°

(ii) (1 – tan² 45°)/(1 + tan² 45°)

1. tan 45° = 1, so tan² 45° = 1.
2. (1 − 1)/(1 + 1) = 0/2 = 0.
3. tan 90° is not defined, and 1 and sin 45° = 1/√2 are not 0, so only (D) matches.

Answer(D) 0

(iii) sin 2A = 2 sin A is true when A =

1. A = 0°: sin 0° = 0 and 2 sin 0° = 0. Equal.
2. A = 30°: sin 60° = √3/2, but 2 sin 30° = 1. Not equal.
3. A = 45°: sin 90° = 1, but 2 sin 45° = 2/√2 = √2. Not equal.
4. A = 60°: sin 120° is not in our table, but 2 sin 60° = √3 is more than 1, and a sine is never more than 1. Not equal.
5. So only A = 0° works.

Answer(A) 0°

(iv) 2 tan 30°/(1 – tan² 30°)

1. 1 − tan² 30° = 1 − 1/3 = 2/3.
2. (2/√3) ÷ (2/3) = (2/√3) × (3/2) = 3/√3 = √3.
3. tan 60° = √3, while cos 60° = 1/2, sin 60° = √3/2 and sin 30° = 1/2, so only (C) matches.

Answer(C) tan 60°

Watch this explained “All thirty cells”, 8:14 into The extreme cases at 0° and 90°, and the ratios that stop being defined

Open NCERT p. 127Matches NCERT’s answer

1. tan (A+B) = √3 is the table value for 60°, and the condition 0° < A+B ≤ 90° makes 60° the only choice, so A + B = 60°.
2. tan (A−B) = 1/√3 is the table value for 30°, so A − B = 30°.
3. Adding the two equations: 2A = 90°, so A = 45°.
4. Substituting back: B = 60° − 45° = 15°.
5. Check: A = 45° > B = 15°, as required.

AnswerA = 45°, B = 15°

Watch this explained “Backwards: an angle from two sides”, 11:48 into Squeezing 30°, 45° and 60° out of two special triangles

Open NCERT p. 127Matches NCERT’s answer

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

1. Take A = B = 30°: sin(A+B) = sin 60° = √3/2, but sin A + sin B = 1/2 + 1/2 = 1.
2. √3/2 is not 1, so the claim fails on this example.

AnswerFalse

(ii) The value of sin θ increases as θ increases.

1. Reading the table from 0° to 90°, the sine values 0, 1/2, 1/√2, √3/2, 1 keep going up.

AnswerTrue

(iii) The value of cos θ increases as θ increases.

1. Reading the table from 0° to 90°, the cosine values 1, √3/2, 1/√2, 1/2, 0 keep going down.

AnswerFalse

(iv) sin θ = cos θ for all values of θ.

1. sin θ and cos θ are equal only at θ = 45°; at 30°, sin 30° = 1/2 but cos 30° = √3/2, which are different.

AnswerFalse

(v) cot A is not defined for A = 0°.

1. cot A = cos A / sin A, and sin 0° = 0, so cot 0° needs a division by zero, which has no value.

AnswerTrue

Watch this explained “Four claims, four verdicts”, 11:29 into The extreme cases at 0° and 90°, and the ratios that stop being defined