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Mathematics Question on Inverse Trigonometric Functions

tan⁡[12sin⁡−1(2x1+x2)+12cos⁡−1(1−x21+x2)]=\tan\left[\frac{1}{2} \sin^{-1} \left(\frac{2x}{1+x^{2}}\right) + \frac{1}{2} \cos^{-1} \left(\frac{1-x^{2}}{1+x^{2}}\right)\right] =

A

∞\infty

B

1

C

2x1−x2 \frac{2x}{1 - x^2}

D

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

Answer

2x1−x2 \frac{2x}{1 - x^2}

Explanation

Solution

tan⁡[12sin⁡−1(2x1+x2)+12cos⁡−1(1−x21+x2)]\tan\left[\frac{1}{2} \sin^{-1} \left(\frac{2x}{1+x^{2}}\right) + \frac{1}{2} \cos^{-1} \left(\frac{1-x^{2}}{1+x^{2}}\right)\right]
Putting x=tan⁡θ⇒θ=tan⁡−1xx = \tan \theta \Rightarrow \theta = \tan^{-1} x
=tan⁡[12sin⁡−1(2tan⁡θ1+tan⁡2θ)+12cos⁡−1(1−tan⁡2θ1+tan⁡2θ)]= \tan \left[\frac{1}{2} \sin ^{-1} \left(\frac{2\tan \theta}{1+\tan^{2} \theta }\right) + \frac{1}{2} \cos ^{-1} \left(\frac{1-\tan ^{2}\theta}{1+\tan ^{2} \theta}\right)\right]
=tan⁡(12sin⁡−1(sin⁡2θ)+12cos⁡−1(cos⁡2θ))=\tan \left(\frac{1}{2}\sin ^{-1} \left(\sin 2 \theta\right) + \frac{1}{2} \cos^{-1} \left(\cos 2\theta\right) \right)
=tan⁡(2θ2+2θ2)=tan⁡(2θ)=tan⁡(2tan⁡−1x)= \tan \left(\frac{2\theta}{2} + \frac{2\theta}{2}\right) =\tan \left(2\theta\right)=\tan \left(2 \tan^{-1} x\right)
=tan⁡(tan⁡−1(2x1−x2))=2x1−x2= \tan \left(\tan^{-1} \left(\frac{2x}{1-x^{2}}\right) \right) =\frac{2x}{1-x^{2}}