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Mathematics Question on Properties of Inverse Trigonometric Functions

The value of 2 tan−1(cosec tan−1x−tan cot−1x))2 \,tan^{-1} (cosec \,tan^{-1} x - tan \,cot^{-1} x)) is

A

tan−1 xtan^{-1}\, x

B

tan xtan \, x

C

cot xcot\, x

D

cosec−1 xcosec^{-1}\, x

Answer

tan−1 xtan^{-1}\, x

Explanation

Solution

2tan⁡−1(cosec⁡tan⁡−1x−tan⁡cot⁡−1x)2 \tan ^{-1}\left(\operatorname{cosec} \tan ^{-1} x-\tan \cot ^{-1} x\right)
= 2 \tan ^{-1}\left[\operatorname{cosec}\left\\{\operatorname{cosec}^{-1} \frac{\sqrt{1+x^{2}}}{x}\right\\}\right.
−tan⁡tan⁡−112]-\tan\\{ \tan^{-1} \frac{1}{2} \\}]
=2tan⁡−1[1+x2x−1x]= 2 \tan ^{-1}\left[\frac{\sqrt{1+x^{2}}}{x}-\frac{1}{x}\right]
= 2 \tan ^{-1}\left\\{\frac{\sqrt{1+x^{2}-1}}{x}\right\\}
= 2 \tan ^{-1}\left\\{\frac{\sec \theta-1}{\tan \theta}\right\\} \,\,(\text { put } x=\tan \theta)
= 2 \tan ^{-1}\left\\{\frac{1-\cos \theta}{\sin \theta}\right\\}
=2 \tan ^{-1}\left\\{\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}}\right\\}
=2tan⁡−1tan⁡θ2=2 \tan ^{-1} \tan \frac{\theta}{2}
=2⋅θ2=θ=tan⁡−1x=2 \cdot \frac{\theta}{2}=\theta=\tan ^{-1} x