Question

If $\frac { 1 } { 2 }$ sin^–1 $\left[ \frac { 3 \sin 2 \theta } { 5 + 4 \cos 2 \theta } \right]$ = tan^–1 x then x =

• A. tan 3q
• B. 3 tan q
• C. $\frac { 1 } { 3 }$ tan q
• D. 3 cot q

Answer 1

Answer: (C)

$\frac { 1 } { 3 }$ tan q

Answer 2

$\frac { 3 \sin 2 \theta } { 5 + 4 \cos 2 \theta }$= $\frac { 3 \times 2 \tan \theta / \left( 1 + \tan ^ { 2 } \theta \right) } { 5 + 4 \frac { \left( 1 - \tan ^ { 2 } \theta \right) } { 1 + \tan ^ { 2 } \theta } }$ =$\frac { 6 \tan \theta } { 9 + \tan ^ { 2 } \theta }$

Put tan q = 3 tan f Ž $\frac { 18 \tan \phi } { 9 + 9 \tan ^ { 2 } \phi }$ = $\frac { 2 \tan \phi } { 1 + \tan ^ { 2 } \phi }$ = sin 2 f

Hence $\frac { 1 } { 2 }$. sin^–1 (sin 2f) = tan^–1 x ; f = tan^–1 x

tan^–1 $\left( \frac { 1 } { 3 } \tan \theta \right)$ = tan^–1 x ; x = $\frac { 1 } { 3 }$ tan q