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Mathematics Question on Inverse Trigonometric Functions

If x≠nπ, x≠ (2n+1)π2.n∈Z,x\neq n \pi ,\, x \neq\,(2n+1)\frac {\pi}{2}.n\in Z, then Sin−1(Cosx)+Cos−1(Sinx)Tan−1(Cotx)+Cot−1(Tanx)\frac {Sin^{-1}(Cos x) + Cos^{-1}(Sin x)}{Tan ^{-1}(Cot x)+ Cot^{-1}(Tan x)} =

A

π6\frac {\pi} {6}

B

π3\frac {\pi} {3}

C

π2\frac {\pi} {2}

D

π4\frac {\pi} {4}

Answer

π4\frac {\pi} {4}

Explanation

Solution

x≠nπ,x≠(2n+1)π2,n∈Zx \neq n \pi, x \neq(2 n+1) \frac{\pi}{2}, n \in Z
Then, sin⁡−1(cos⁡x)+cos⁡−1(sin⁡x)tan⁡−1(cot⁡x)+cot⁡−1(tan⁡x)\frac{\sin ^{-1}(\cos x)+\cos ^{-1}(\sin x)}{\tan ^{-1}(\cot x)+\cot ^{-1}(\tan x)}
=\frac{\sin ^{-1}\left\\{\sin \left(\frac{\pi}{2}-x\right)\right\\}+\cos ^{-1}\left\\{\cos \left(\frac{\pi}{2}-x\right)\right\\}}{\tan ^{-1}\left\\{\tan \left(\frac{\pi}{2}-x\right)\right\\}+\cot ^{-1}\left\\{\cot \left(\frac{\pi}{2}-x\right)\right\\}}
=(π2−x)+(π2−x)(π2−x)+(π2−x)=(π−2xπ−2x)=1=\frac{\left(\frac{\pi}{2}-x\right)+\left(\frac{\pi}{2}-x\right)}{\left(\frac{\pi}{2}-x\right)+\left(\frac{\pi}{2}-x\right)}=\left(\frac{\pi-2 x}{\pi-2 x}\right)=1
But from the option we take x=π4x=\frac{\pi}{4}
=sin⁡−1(cos⁡π4)+cos⁡−1(sin⁡π4)tan⁡−1(cot⁡π4)+cot⁡−1(tan⁡π4)=\frac{\sin ^{-1}\left(\cos \frac{\pi}{4}\right)+\cos ^{-1}\left(\sin \frac{\pi}{4}\right)}{\tan ^{-1}\left(\cot \frac{\pi}{4}\right)+\cot ^{-1}\left(\tan \frac{\pi}{4}\right)}
=sin⁡−1(12)+cos⁡−1(12)tan⁡−1(1)+cot⁡−1(1)=\frac{\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)+\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)}{\tan ^{-1}(1)+\cot ^{-1}(1)}
{sin⁡−1x+cos⁡−1x=π2 ∵ tan⁡−1x+cot−1x=π2}\begin{Bmatrix}\sin^{-1}x +\cos^{-1}&x=\frac{\pi}{2}\\\ \because&\\\ \tan^{-1} x +cot^{-1}&x=\frac{\pi}{2}\end{Bmatrix}
=π/2π/2=1=\frac{\pi / 2}{\pi / 2}=1