Q87 / 200 · Manipal (MET) 2010 · Mathematics
The value of
$\sin ^{-1}\left[\operatorname { c o s } \left(\sin ^{-1} \sqrt{\frac{2-\sqrt{3}}{4}}\right.\right.\left.\left.+\cos ^{-1} \frac{\sqrt{12}}{4}+\sec ^{-1} \sqrt{2}\right)\right]$ is
- Correct answerA0
- B$\frac{\pi}{4}$
- C$\frac{\pi}{6}$
- D$\frac{\pi}{2}$
Solution
We have,
$\sin ^{-1}\left[\cos \left(\sin ^{-1} \sqrt{\left(\frac{2-\sqrt{3}}{4}\right)}+\cos ^{-1} \frac{\sqrt{12}}{4}+\sec ^{-1} \sqrt{2}\right)\right]$
$=\sin ^{-1}\left[\cos \left(\sin ^{-1}\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)+\cos ^{-1} \frac{\sqrt{3}}{2}+\cos ^{-1} \frac{1}{\sqrt{2}}\right)\right]$
$\left[\because\left(\frac{2-\sqrt{3}}{4}\right)^{1 / 2}=\frac{\sqrt{3}-1}{2 \sqrt{2}}\right]$
$=\sin ^{-1}\left[\cos \left(15^{\circ}+30^{\circ}+45^{\circ}\right)\right]$
$=\sin ^{-1}\left[\cos 90^{\circ}\right] \quad\left(\because \sin 15^{\circ}=\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)$
$\begin{aligned} & =\sin ^{-1} 0 \\ & =0\end{aligned}$
Hence, the correct answer is option: 'A'.