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Question: \(\tan ^ { - 1 } \left[ \frac { \cos x } { 1 + \sin x } \right] =\)...

tan⁡−1[cos⁡x1+sin⁡x]=\tan ^ { - 1 } \left[ \frac { \cos x } { 1 + \sin x } \right] =

A

π4−x2\frac { \pi } { 4 } - \frac { x } { 2 }

B

π4+x2\frac { \pi } { 4 } + \frac { x } { 2 }

C

x2\frac { x } { 2 }

D

π4−x\frac { \pi } { 4 } - x

Answer

π4−x2\frac { \pi } { 4 } - \frac { x } { 2 }

Explanation

Solution

tan⁡−1[cos⁡x1+sin⁡x]=tan⁡−1[sin⁡(π/2−x)1+cos⁡(π/2−x)]\tan ^ { - 1 } \left[ \frac { \cos x } { 1 + \sin x } \right] = \tan ^ { - 1 } \left[ \frac { \sin ( \pi / 2 - x ) } { 1 + \cos ( \pi / 2 - x ) } \right]

=tan⁡−1[2sin⁡(π/4−x/2)cos⁡(π/4−x/2)2cos⁡2(π/4−x/2)]= \tan ^ { - 1 } \left[ \frac { 2 \sin ( \pi / 4 - x / 2 ) \cos ( \pi / 4 - x / 2 ) } { 2 \cos ^ { 2 } ( \pi / 4 - x / 2 ) } \right]

=tan⁡−1tan⁡(π4−x2)=π4−x2= \tan ^ { - 1 } \tan \left( \frac { \pi } { 4 } - \frac { x } { 2 } \right) = \frac { \pi } { 4 } - \frac { x } { 2 }.