Question:
If $\sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$, then $x=$
(a) $\frac{1}{2}$
(b) $\frac{\sqrt{3}}{2}$
(c) $-\frac{1}{2}$
(d) none of these
Solution:
(b) $\frac{\sqrt{3}}{2}$
We know that $\sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}$.
$\therefore \sin ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$
$\Rightarrow \frac{\pi}{2}-\cos ^{-1} x-\cos ^{-1} x=\frac{\pi}{6}$
$\Rightarrow-2 \cos ^{-1} x=\frac{\pi}{6}-\frac{\pi}{2}$
$\Rightarrow-2 \cos ^{-1} x=-\frac{\pi}{3}$
$\Rightarrow \cos ^{-1} x=\frac{\pi}{6}$
$\Rightarrow x=\cos \frac{\pi}{6}$
$\Rightarrow x=\frac{\sqrt{3}}{2}$
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