(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 = 2 sin 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°

NCERT Class 10 Chapter 8 Introduction to Trigonometry

Page No. 187

Exercise 8.2

Question No. 2

(i) (2 tan 30°)/(1 – tan²30°)

Putting the value of each trigonometric ratios, we get

(2(1/√3))/(1+(1/√3)²) = (2/√3)/(1 + 1/3) = (2/√3)/(4/3) = (6)/(4√3) = √3/2

We know that sin 60° =Y hence the option (A) is correct.

(ii)(1 – tan² 45°)/(1 + tan² 45°)

Putting the value of each trigonometric ratios, we get

(1-(1)²)(1+(1)²) = (1-1)/(1+1) = 0/2 = 0

Hence, the option (D) is correct.

(iii) sin 2A = 2 sin A

We know that sin 0 = 0, hence, the option (A) is correct.

(iv) (2 tan 30°)/(1 – tan²30°)

Putting the value of each trigonometric ratios, we get

(2(1/√3))/(1-(1/√3)²) = (1/√3)/(1 – 1/3) = (2/√3)/(2/3) = 3/√3 = √3

We know that tan 60° = √3, hence, the option (C) is correct.

Here is the video explanation of the above question😁🤗