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Answer :
(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
\( \frac{2tan^230 °}{1- tan^230 °} \)
\( = \ \frac{2 × \frac{1}{ \sqrt{3}}}{1 - ( \frac{1}{ \sqrt{3}})^2} \)
\( = \ \frac{ \frac{2}{ \sqrt{3}}}{1 - \frac{1}{3}} \)
\( = \ \frac{2}{ \sqrt{3}} × \frac{3}{3 - 1} \ \)
\( = \ \frac{2}{ \sqrt{3}} × \frac{3}{2} \)
\( = \ \sqrt{3} \ = \ tan60° \)