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Question

Question: If \(A = \tan ^ { - 1 } x\) , then \(\sin 2 A =\)...

If A=tan⁡−1xA = \tan ^ { - 1 } x , then sin⁡2A=\sin 2 A =

A

2x1−x2\frac { 2 x } { \sqrt { 1 - x ^ { 2 } } }

B

2x1−x2\frac { 2 x } { 1 - x ^ { 2 } }

C

2x1+x2\frac { 2 x } { 1 + x ^ { 2 } }

D

None of these

Answer

2x1+x2\frac { 2 x } { 1 + x ^ { 2 } }

Explanation

Solution

Given that A=tan⁡−1xA = \tan ^ { - 1 } x

Now x=tan⁡A⇒sin⁡2A=2tan⁡A1+tan⁡2A=2x1+x2x = \tan A \Rightarrow \sin 2 A = \frac { 2 \tan A } { 1 + \tan ^ { 2 } A } = \frac { 2 x } { 1 + x ^ { 2 } } .