# Choose the correct option and justify your choice:

(i) 2 tan 30°/1 + tan^{2} 30° =

(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°

(ii) 1 - tan^{2} 45°/1 + tan^{2} 45° =

(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 - tan^{2} 30° =

(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°

**Solution:**

We will use the six basic trigonometric ratios and trigonometric table to solve the problem.

(i) 2 tan 30°/1 + tan^{2} 30°

By substituting the values of given trigonometric ratios in the above equation, we get

= 2 × (1/√3)/1 + (1/√3)^{2}

= 2 × (1/√3)/(1 + 1/3)

= (2/√3) / (4/3)

= 6/4√3

= √3/2

Out of the given options only sin 60° = √3/2. Hence, option (A) is correct.

(ii) 1 - tan² 45°/1 + tan² 45°

By substituting the values of given trigonometric ratios for tan 45°.

= 1 - (1)²/1 + (1)²

= (1 - 1)/(1 + 1)

= 0/2

= 0

Hence, option (D) is correct.

(iii) sin 2A = 2 sin A

By substituting A = 0°, 30°, 45° and 60°, we get

For A = 0°,

sin 2A = sin (2 × 0°)

= sin 0°

= 0

2 sin A = 2 × sin 0°

= 2 × 0

= 0

So, sin 2A = 2 sin A, when A = 0°

For A = 30°,

sin 2A = sin (2 × 30)°

= sin 60°

= √3/2

2 sin A = 2 × sin 30°

= 2 × 1/2

= 1

sin 2A ≠ 2 sin A, when A = 30°

For A = 45°

sin 2A = sin (2 × 45)°

= sin 90°

= 1

2 sin A = 2 × sin 45°

= 2 × 1/√2

= √2

So, sin 2A ≠ 2sin A, when A = 45°

For A = 60°

sin 2A = sin 2 × 60°

= sin 120°

= sin (180° - 60^{°})

= sin 60°

= √3/2

2sin A = 2 × sin 60°

= 2 × √3/2

= √3

So, sin 2A ≠ 2 sin A, when A = 60°

Hence, Option (A) is correct.

(iv) 2 tan 30°/1 - tan^{2} 30°

By substituting the values of given trigonometric ratios for tan 30°, we get

= 2 × (1/√3) / 1 - (1/√3)^{2}

= (2/√3) / (1 - 1/3)

= (2/√3) / (2/3)

= √3

Out of the given option only tan 60° = √3.

Hence, option (C) is correct.

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 8

**Video Solution:**

## Choose the correct option and justify your choice: (i) 2 tan 30°/1 + tan² 30° (A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30° (ii) 1 - tan² 45°/1 + tan² 45° (A) tan 90° (B) 1 (C) sin 45° (D) 0 (iii) sin 2A = 2sin A is true when A = (A) 0° (B) 30° (C) 45° (D) 60° (iv) 2 tan 30°/1 - tan² 30° (A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°

Maths NCERT Solutions Class 10 Chapter 8 Exercise 8.2 Question 2

**Summary:**

The correct option for the following are (i) 2 tan 30°/1 + tan^{2} 30° - option (A) is correct, (ii) 1 - tan² 45°/1 + tan² 45° - option (D) is correct, (iii) in 2A = 2sin A - option (A) is correct, (iv) 2 tan 30°/1 - tan^{2} 30° - option (C) is correct.

**☛ Related Questions:**

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- State whether the following are true or false. Justify your answer.(i) sin (A + B) = sin A + sin B.(ii) The value of sin θ increases as θ.(iii) The value of cos θ increases as θ.(iv) sin θ = cos θ for all values of θ.(v) cot A is not defined for A = 0°.
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