Question

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

• A. 0
• B. 1
• C. -1
• D. $\pi$

Answer 1

Answer: (B)

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