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Question: \(\tan ^ { - 1 } x + \cot ^ { - 1 } ( x + 1 ) =\)...

tan⁡−1x+cot⁡−1(x+1)=\tan ^ { - 1 } x + \cot ^ { - 1 } ( x + 1 ) =

A

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

B

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

C

tan⁡−1(x+1)\tan ^ { - 1 } ( x + 1 )

D

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

Answer

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

Explanation

Solution

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

=tan⁡−1[x+1x+11−xx+1]=tan⁡−1(x2+x+1)= \tan ^ { - 1 } \left[ \frac { x + \frac { 1 } { x + 1 } } { 1 - \frac { x } { x + 1 } } \right] = \tan ^ { - 1 } \left( x ^ { 2 } + x + 1 \right) .