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Question

Mathematics Question on Inverse Trigonometric Functions

Find the value of sin⁡ (2 tan⁡−1x), ∣x∣≤1\sin \,(2\,{{\tan }^{-1}}x),\,|x|\le 1

A

1/x1/x

B

xx

C

1/x21/{{x}^{2}}

D

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

Answer

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

Explanation

Solution

sin⁡(2tan⁡−1x)=sin⁡(sin⁡−12x1+x2)\sin (2{{\tan }^{-1}}x)=\sin \left( {{\sin }^{-1}}\frac{2x}{1+{{x}^{2}}} \right)
=2x1+x2,−1≤x≤1=\frac{2x}{1+{{x}^{2}}},-1\le x\le 1