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Question: \(\tan ^ { - 1 } \left( \frac { \sqrt { 1 + x ^ { 2 } } - 1 } { x } \right) =\)...

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

A

tan⁡−1x\tan ^ { - 1 } x

B

12tan⁡−1x\frac { 1 } { 2 } \tan ^ { - 1 } x

C

2tan⁡−1x2 \tan ^ { - 1 } x

D

None of these

Answer

12tan⁡−1x\frac { 1 } { 2 } \tan ^ { - 1 } x

Explanation

Solution

tan⁡−1(1+x2−1x)=tan⁡−1[1+tan⁡2θ−1tan⁡θ]\tan ^ { - 1 } \left( \frac { \sqrt { 1 + x ^ { 2 } } - 1 } { x } \right) = \tan ^ { - 1 } \left[ \frac { \sqrt { 1 + \tan ^ { 2 } \theta } - 1 } { \tan \theta } \right]

(Putting x=tan⁡θ)x = \tan \theta )

=tan⁡−1[sec⁡θ−1tan⁡θ]=tan⁡−1[1−cos⁡θsin⁡θ]= \tan ^ { - 1 } \left[ \frac { \sec \theta - 1 } { \tan \theta } \right] = \tan ^ { - 1 } \left[ \frac { 1 - \cos \theta } { \sin \theta } \right]

=tan⁡−1tan⁡θ2=θ2=12tan⁡−1x= \tan ^ { - 1 } \tan \frac { \theta } { 2 } = \frac { \theta } { 2 } = \frac { 1 } { 2 } \tan ^ { - 1 } x.