A) \[\frac{\pi }{2}\]
B) \[\frac{\pi }{6}\]
C) \[\frac{\pi }{4}\]
D) \[\frac{\pi }{3}\]
Correct Answer: C
Solution :
\[x\ne n\pi ,x\ne (2n+1)\frac{\pi }{2},n\in Z\] Then, \[\frac{{{\sin }^{-1}}(\cos x)+{{\cos }^{-1}}(\sin x)}{{{\tan }^{-1}}(\cos x)+{{\cot }^{-1}}(\tan x)}\] \[=\frac{{{\sin }^{-1}}\left\{ \sin \left( \frac{\pi }{2}-x \right) \right\}+{{\cos }^{-1}}\left\{ \cos \left( \frac{\pi }{2}-x \right) \right\}}{{{\tan }^{-1}}\left\{ \tan \left( \frac{\pi }{2}-x \right) \right\}+{{\cot }^{-1}}\left\{ \cot \left( \frac{\pi }{2}-x \right) \right\}}\] \[=\frac{\left( \frac{\pi }{2}-x \right)+\left( \frac{\pi }{2}-x \right)}{\left( \frac{\pi }{2}-x \right)+\left( \frac{\pi }{2}-x \right)}=\left( \frac{\pi -2x}{\pi -2x} \right)=1\] But from the option we take \[x=\frac{\pi }{4}\] \[=\frac{{{\sin }^{-1}}\left( \cos \frac{\pi }{4} \right)+{{\cos }^{-1}}\left( \sin \frac{\pi }{4} \right)}{{{\tan }^{-1}}\left( \cot \frac{\pi }{4} \right)+{{\cot }^{-1}}\left( \tan \frac{\pi }{4} \right)}\] \[=\frac{{{\sin }^{-1}}\left( \frac{1}{\sqrt{2}} \right)+{{\cos }^{-1}}\left( \frac{1}{\sqrt{2}} \right)}{{{\tan }^{-1}}(1)+co{{t}^{-1}}(1)}\] \[\left\{ \because \,\,\begin{matrix} {{\sin }^{-1}}x+{{\cos }^{-1}}x=\frac{\pi }{2} \\ {{\tan }^{-1}}x+{{\cot }^{-1}}x=\frac{\pi }{2} \\ \end{matrix} \right\}\] \[=\frac{\pi /2}{\pi /2}=1\]
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