Q67 / 180 · KCET 2010 · Mathematics
If $\mathrm{x} \neq \mathrm{n} \pi, \mathrm{x} \neq(2 \mathrm{n}+1) \frac{\pi}{2}, \mathrm{n} \in \mathrm{Z}, \quad$ then $\frac{\sin ^{-1}(\cos x)+\cos ^{-1}(\sin x)}{\tan ^{-1}(\cot x)+\cot ^{-1}(\tan x)}$ is
- A$\frac{\pi}{2}$
- B$\frac{\pi}{6}$
- Correct answerC$\frac{\pi}{4}$
- D$\frac{\pi}{3}$
Solution
$\mathrm{x} \neq \mathrm{n} \pi, \mathrm{x} \neq(2 \mathrm{n}+1) \frac{\pi}{2}, \mathrm{n} \in \mathrm{Z}$
Then, $\quad \frac{\sin ^{-1}(\cos \mathrm{x})+\cos ^{-1}(\sin \mathrm{x})}{\tan ^{-1}(\cot \mathrm{x})+\cot ^{-1}(\tan \mathrm{x})}$
$=\frac{\sin ^{-1}\left\{\sin \left(\frac{\pi}{2}-\mathrm{x}\right)\right\}+\cos ^{-1}\left\{\cos \left(\frac{\pi}{2}-\mathrm{x}\right)\right\}}{\tan ^{-1}\left\{\tan \left(\frac{\pi}{2}-\mathrm{x}\right)\right\}+\cot ^{-1}\left\{\cot \left(\frac{\pi}{2}-\mathrm{x}\right)\right\}}$
$=\frac{\left(\frac{\pi}{2}-\mathrm{x}\right)+\left(\frac{\pi}{2}-\mathrm{x}\right)}{\left(\frac{\pi}{2}-\mathrm{x}\right)+\left(\frac{\pi}{2}-\mathrm{x}\right)}=\left(\frac{\pi-2 \mathrm{x}}{\pi-2 \mathrm{x}}\right)=1$
But from the option we take $x=\frac{\pi}{4}$
$$
\begin{aligned}
&=\frac{\sin ^{-1}\left(\cos \frac{\pi}{4}\right)+\cos ^{-1}\left(\sin \frac{\pi}{4}\right)}{\tan ^{-1}\left(\cot \frac{\pi}{4}\right)+\cot ^{-1}\left(\tan \frac{\pi}{4}\right)} \\
&=\frac{\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)+\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)}{\tan ^{-1}(1)+\cot ^{-1}(1)}\left\{\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right\} \\
&\left.=\frac{\pi / 2}{\pi / 2}=1 \quad \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right\}
\end{aligned}
$$
Hence, the correct answer is option: 'C'.