Question

If $x\ne n\pi ,x\ne (2n+1)\frac{\pi }{2}.n\in Z,$ then $\frac{Si{n}^{-1}(Cosx)+Co{s}^{-1}(Sinx)}{Ta{n}^{-1}(Cotx)+Co{t}^{-1}(Tanx)}$ =

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{4}$

Answer 1

Answer: (D)

$x\ne n\pi ,x\ne (2n+1)\frac{\pi }{2},n\in Z$
Then, $\frac{{\sin }^{-1}(\cos x)+{\cos }^{-1}(\sin x)}{{\tan }^{-1}(\cot x)+{\cot }^{-1}(\tan x)}$
$=\frac{{\sin }^{-1}\left\{\sin \left(\frac{\pi }{2}-x\right)\right\}+{\cos }^{-1}\left\{\cos \left(\frac{\pi }{2}-x\right)\right\}}{{\tan }^{-1}\left\{\tan \left(\frac{\pi }{2}-x\right)\right\}+{\cot }^{-1}\left\{\cot \left(\frac{\pi }{2}-x\right)\right\}}$
$=\frac{\left(\frac{\pi }{2}-x\right)+\left(\frac{\pi }{2}-x\right)}{\left(\frac{\pi }{2}-x\right)+\left(\frac{\pi }{2}-x\right)}=\left(\frac{\pi -2x}{\pi -2x}\right)=1$
But from the option we take $x=\frac{\pi }{4}$
$=\frac{{\sin }^{-1}\left(\cos \frac{\pi }{4}\right)+{\cos }^{-1}\left(\sin \frac{\pi }{4}\right)}{{\tan }^{-1}\left(\cot \frac{\pi }{4}\right)+{\cot }^{-1}\left(\tan \frac{\pi }{4}\right)}$
$=\frac{{\sin }^{-1}\left(\frac{1}{\sqrt{2}}\right)+{\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)}{{\tan }^{-1}(1)+{\cot }^{-1}(1)}$
$\left\{\begin{array}{cc}{\sin }^{-1}x+{\cos }^{-1}& x=\frac{\pi }{2}\\ \because & \\ {\tan }^{-1}x+co{t}^{-1}& x=\frac{\pi }{2}\end{array}\right\}$
$=\frac{\pi /2}{\pi /2}=1$