Choose the correct option and justify :

(i) \( \frac{2tan30°}{1+tan^230°} \) =

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

(ii) \( \frac{1 - tan^245 °}{1+tan^245 °} \) =

(A) tan 90°
(B) 1
(C) sin 45 °
(D) 0

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

(A) 0°
(B) 30°
(C) 45°
(D) 60°

(iv) \( \frac{2tan^230 °}{1- tan^230 °} \) =

(A) cos 60°
(B) sin 60°
(C) tan 60 °
(D) none of these

(i) (A), as
\( \frac{2tan30°}{1+tan^230°} \)
\( = \ \frac{2 × \frac{1}{ \sqrt{3}}}{1+( \frac{1}{ \sqrt{3}})^2} \ \)
\( = \ \frac{ \frac{2}{ \sqrt{3}}}{1 + \frac{1}{3}} \)
\( = \ \frac{2}{ \sqrt{3}} × \frac{3}{4} \)
\( = \ \frac{ \sqrt{3}}{2} \ = \ sin60 ° \)


(ii) (D), as
\( \frac{1- tan^245 °}{1+tan^245 °} \)
\( =\ \frac{1- 1}{1+1} \ = \ 0 \)


(iii) (A), as
when A = 0, sin 2 A = sin 0 = 0

and, 2 sinA = 2 sin 0 = 2 × 0 = 0
=> sin 2A = 2sinA, when A = 0

(iv) (C), as
\( \frac{2tan^230 °}{1- tan^230 °} \)
\( = \ \frac{2 × \frac{1}{ \sqrt{3}}}{1 - ( \frac{1}{ \sqrt{3}})^2} \)
\( = \ \frac{ \frac{2}{ \sqrt{3}}}{1 - \frac{1}{3}} \)
\( = \ \frac{2}{ \sqrt{3}} × \frac{3}{3 - 1} \ \)
\( = \ \frac{2}{ \sqrt{3}} × \frac{3}{2} \)
\( = \ \sqrt{3} \ = \ tan60° \)