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Question: A solution of the equation \(\tan ^ { - 1 } ( 1 + x )\) \(+ \tan ^ { - 1 } ( 1 - x )\) \(= \frac ...

A solution of the equation tan⁡−1(1+x)\tan ^ { - 1 } ( 1 + x ) +tan⁡−1(1−x)+ \tan ^ { - 1 } ( 1 - x ) =π2= \frac { \pi } { 2 } is .

A

x=1x = 1

B

x=−1x = - 1

C

x=0x = 0

D

x=πx = \pi

Answer

x=0x = 0

Explanation

Solution

tan⁡−1(1+x)+tan⁡−1(1−x)=π2\tan ^ { - 1 } ( 1 + x ) + \tan ^ { - 1 } ( 1 - x ) = \frac { \pi } { 2 }

⇒ tan⁡−1(1+x)=π2−tan⁡−1(1−x)\tan ^ { - 1 } ( 1 + x ) = \frac { \pi } { 2 } - \tan ^ { - 1 } ( 1 - x )

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

⇒ tan⁡−1(1+x)=tan⁡−1(11−x)\tan ^ { - 1 } ( 1 + x ) = \tan ^ { - 1 } \left( \frac { 1 } { 1 - x } \right)

⇒ 1+x=11−x⇒1−x2=1⇒x=01 + x = \frac { 1 } { 1 - x } \Rightarrow 1 - x ^ { 2 } = 1 \Rightarrow x = 0.