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Question: \(\cos ^ { - 1 } \left( \frac { 3 + 5 \cos x } { 5 + 3 \cos x } \right)\) is equal to....

cos⁡−1(3+5cos⁡x5+3cos⁡x)\cos ^ { - 1 } \left( \frac { 3 + 5 \cos x } { 5 + 3 \cos x } \right) is equal to.

A

tan⁡−1(12tan⁡x2)\tan ^ { - 1 } \left( \frac { 1 } { 2 } \tan \frac { x } { 2 } \right)

B

2tan⁡−1(2tan⁡x2)2 \tan ^ { - 1 } \left( 2 \tan \frac { x } { 2 } \right)

C

12tan⁡−1(2tan⁡x2)\frac { 1 } { 2 } \tan ^ { - 1 } \left( 2 \tan \frac { x } { 2 } \right)

D

2tan⁡−1(12tan⁡x2)2 \tan ^ { - 1 } \left( \frac { 1 } { 2 } \tan \frac { x } { 2 } \right)

(5) tan⁡−1(tan⁡x2)\tan ^ { - 1 } \left( \tan \frac { x } { 2 } \right)

Answer

2tan⁡−1(12tan⁡x2)2 \tan ^ { - 1 } \left( \frac { 1 } { 2 } \tan \frac { x } { 2 } \right)

55 tan⁡−1(tan⁡x2)\tan ^ { - 1 } \left( \tan \frac { x } { 2 } \right)

Explanation

Solution

We take x=π2x = \frac { \pi } { 2 } then cos⁡x=0\cos x = 0

cos⁡−1(3+5cos⁡x5+3cos⁡x)=cos⁡−1(35)\cos ^ { - 1 } \left( \frac { 3 + 5 \cos x } { 5 + 3 \cos x } \right) = \cos ^ { - 1 } \left( \frac { 3 } { 5 } \right) =tan⁡−1(43)= \tan ^ { - 1 } \left( \frac { 4 } { 3 } \right)

Put 2tan⁡−1(12tan⁡x2)2 \tan ^ { - 1 } \left( \frac { 1 } { 2 } \tan \frac { x } { 2 } \right)

we get 2tan⁡−1(12tan⁡π4)2 \tan ^ { - 1 } \left( \frac { 1 } { 2 } \tan \frac { \pi } { 4 } \right)

=tan⁡−1(43)= \tan ^ { - 1 } \left( \frac { 4 } { 3 } \right).