Question
If $${\sin ^{ - 1}}x - {\cos ^{ - 1}}x = \frac{\pi }{6}$$ then $$x$$ is
A.
$$\frac{1}{2}$$
B.
$$\frac{{\sqrt 3 }}{2}$$
C.
$$ - \frac{1}{2}$$
D.
None of these
Answer :
$$\frac{{\sqrt 3 }}{2}$$
Solution :
$$\eqalign{
& \left( {\frac{\pi }{2} - {{\cos }^{ - 1}}x} \right) - {\cos ^{ - 1}}x = \frac{\pi }{6}\,\,{\text{or, }}2{\cos ^{ - 1}}x = \frac{\pi }{2} - \frac{\pi }{6} = \frac{\pi }{3}\,\,{\text{or, }}{\cos ^{ - 1}}x = \frac{\pi }{6} \cr
& \therefore \,\,x = \frac{{\sqrt 3 }}{2}. \cr} $$