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Q.
If $\operatorname{cosec}^{-1} x+\sec ^{-1} x^2+\frac{\pi}{2}=0$ is satisfied by $x=\alpha$ then $\sin ^{-1} \alpha-\cos ^{-1} \alpha$ is equal to $
Inverse Trigonometric Functions
Solution:
$\Theta \operatorname{cosec}^{-1} x +\sec ^{-1} x ^2+\frac{\pi}{2}=0 $
$\Rightarrow \operatorname{cosec}^{-1} x =-\frac{\pi}{2} \text { and } \sec ^{-1} x ^2=0$
$\Rightarrow x =-1 \text { and } x ^2=1 \Rightarrow x = \pm 1 \\
\therefore \alpha=-1$
$\therefore \sin ^{-1} \alpha-\cos ^{-1} \alpha=-\frac{\pi}{2}-\pi=-\frac{3 \pi}{2}
$