Question
If $${\cos ^{ - 1}}x > {\sin ^{ - 1}}x$$ then
A.
$$x < 0$$
B.
$$ - 1 < x < 0$$
C.
$$0 \leqslant x < \frac{1}{{\sqrt 2 }}$$
D.
$$- 1 \leqslant x < \frac{1}{{\sqrt 2 }}$$
Answer :
$$- 1 \leqslant x < \frac{1}{{\sqrt 2 }}$$
Solution :
For $${\cos ^{ - 1}}x,{\sin ^{ - 1}}x$$ to be real, $$ - 1 \leqslant x \leqslant 1.$$
But $${\cos ^{ - 1}}x > {\sin ^{ - 1}}x$$
$$\eqalign{
& \Rightarrow \,\,2{\cos ^{ - 1}}x > \frac{\pi }{2} \cr
& \Rightarrow \,\,{\cos ^{ - 1}}x > \frac{\pi }{4} \cr
& \therefore \,\,\cos \left( {{{\cos }^{ - 1}}x} \right) < \cos \frac{\pi }{4}\,\,{\text{or, }}x < \frac{1}{{\sqrt 2 }}. \cr} $$
The common values are $$ - 1 \leqslant x < \frac{1}{{\sqrt 2 }}.$$