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Question

Mathematics Question on Inverse Trigonometric Functions

Which of the following is not the value of 2 tan⁡−1x2\,\tan^{-1}x ?

A

tan⁡−12x1+x2\tan^{-1}\frac{2x}{1+x^2}

B

tan⁡−12x1−x2\tan^{-1}\frac{2x}{1-x^2}

C

cos⁡−11−x21+x2\cos^{-1}\frac{1-x^2}{1+x^2}

D

sin⁡−12x1+x2\sin^{-1}\frac{2x}{1+x^2}

Answer

tan⁡−12x1+x2\tan^{-1}\frac{2x}{1+x^2}

Explanation

Solution

Since 2 tan θ1−tan2θ=tan 2 θ \frac{2\,tan\,\theta}{1-tan^{2}\theta} = tan \,2\,\theta 1−tan2θ1+tan2θ=cos 2 θ,2 tan θ1+tan θ=sin 2 θ\frac{ 1-tan ^{2}\theta}{1+tan^{2}\theta} = cos \,2\,\theta,\frac{ 2\,tan \,\theta}{1+tan\,\theta} = sin \,2\,\theta ∴(b)=2 θ,(c)=2 θ,(d)=2 θ=2 tan−1x\therefore \left(b\right)= 2\,\theta, \left(c\right)=2\,\theta, \left(d\right)= 2\,\theta = 2\,tan^{-1}x Hence tan−12x1+x2≠2 tan−1xtan^{-1} \frac{2x}{1+x^{2}} \ne 2\,tan^{-1} x