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Question

Mathematics Question on Inverse Trigonometric Functions

Write the function in the simplest form: tan⁡−11+x2−1x,x≠0.\tan^{-1}\frac{\sqrt1+x^2-1}{x},x\neq0.

Answer

tan⁡−11+x2−1x\tan^{-1}\frac{\sqrt1+x^2-1}{x},

put x=\tan\theta$$\Rightarrow θ\theta= tan⁡−1x\tan^{-1}x

∴tan⁡−11+x2−1x\therefore \tan^{-1}\frac{\sqrt1+x^2-1}{x} =tan⁡−11+tan⁡2θ−1tan⁡θ\tan^{-1}\frac{\sqrt1+\tan2\theta-1}{\tan\theta}

= tan⁡−1(sec⁡θ−1)tan⁡θ=tan⁡−1(1−cos⁡θsin⁡θ)\tan^{-1}\frac{(\sec\theta-1)}{\tan\theta}=\tan^{-1}(\frac{1-\cos\theta}{\sin\theta})

=tan⁡−12sin⁡2θ22sin⁡θ22cos⁡θ2\tan^{-1}\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}2\cos\frac{\theta}{2}}

=tan⁡−1(tan⁡θ2)=θ2=12tan⁡−1x\tan^{-1}(\frac{\tan\theta}{2})=\frac{\theta}{2}=\frac{1}{2}\tan^{-1}x