Question

Find the interval of x if ${\cos ^{ - 1}}x > {\sin ^{ - 1}}x$
${\text{A}}.$$[ - \infty ,0]$
${\text{B}}.$ $[ - 1,0]$
\[{\text{C}}{\text{.}}\] $[0,\dfrac{1}{{\sqrt 2 }}]$
${\text{D}}.$ $[ - 1,\dfrac{1}{{\sqrt 2 }}]$

Answer 1

Hint: To solve this question use the result, ${\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \dfrac{\pi }{2}$. We can write it as ${\sin ^{ - 1}}x = \dfrac{\pi }{2} - {\cos ^{ - 1}}x$.

Complete step-by-step answer:
Given,
${\cos ^{ - 1}}x > {\sin ^{ - 1}}x - (1)$
Put ${\sin ^{ - 1}}x = \dfrac{\pi }{2} - {\cos ^{ - 1}}x$ in equation (1), we get
\[{\cos ^{ - 1}}x > \dfrac{\pi }{2} - {\cos ^{ - 1}}x\]
$
   \Rightarrow 2{\cos ^{ - 1}}x > \dfrac{\pi }{2} \\
   \Rightarrow {\cos ^{ - 1}}x > \dfrac{\pi }{4} \\
 $
Therefore, $x < \cos \left( {\dfrac{\pi }{4}} \right)$ (Since, $\cos \theta $ is a decreasing function).
$ \Rightarrow x < \dfrac{1}{{\sqrt 2 }}$
Now, we know that the domain of ${\cos ^{ - 1}}$ is [-1,1]
Hence, $x \in \left[ { - 1,\dfrac{1}{{\sqrt 2 }}} \right]$.
Therefore, the correct option is ${\text{D}}.$$[ - 1,\dfrac{1}{{\sqrt 2 }}]$.

Note: Whenever such a type of question appears, then always try to write either sine in terms of cosine or vice- versa, so that the whole equation converts into similar terms, which is easier to solve. In the solution, sine is written in terms of cosine. Make sure that you know the domain of the cosine function. Also, as we know, $\cos \theta $ is a decreasing function, so care must be taken while solving the inequality.