Question

If $x=\alpha$ satisfies the equation $\sin ^{-1} x+\cos ^{-1} x^2+\frac{\pi}{2}=0$, then the value of $\frac{\sec ^{-1} \alpha-\tan ^{-1} \alpha}{\cot ^{-1} \alpha-\operatorname{cosec}^{-1} \alpha}$ is equal to

• A. 0
• B. 1
• C. -1
• D. $\pi$

Answer 1

Answer: (B)

$ \because \sin ^{-1} x+\cos ^{-1} x^2+\frac{\pi}{2}=0 $
$\Rightarrow \frac{\pi}{2}+\sin ^{-1} x=-\cos ^{-1} x^2$
$\Rightarrow \cos \left(\frac{\pi}{2}+\sin ^{-1} x\right)=\cos \left(-\cos ^{-1} x^2\right) $
$\Rightarrow - x = x ^2 $
$\Rightarrow x =0,-1 ( x =0 \text { rejected }) $
$\therefore x =-1=\alpha$
$\therefore \text { Given expression }=\frac{\sec ^{-1}(-1)-\tan ^{-1}(-1)}{\cot ^{-1}(-1)-\operatorname{cosec}^{-1}(-1)}=\frac{\pi-\left(\frac{-\pi}{4}\right)}{\frac{3 \pi}{4}-\left(\frac{-\pi}{2}\right)}=\frac{\frac{5 \pi}{4}}{\frac{5 \pi}{4}}=1 $