Question

If ${\sec }^{-1}\frac{1}{\sqrt{1-x^2}}+{\cot }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)={\sin }^{-1}\left(k\right)$ then k =

• A. $x\sqrt{1-x^2}$
• B. $2x\sqrt{1-x^2}$
• C. $\sqrt{1-x^2}$
• D. $2x$

Answer 1

The correct option is B $2x\sqrt{1-x^2}$

Given

${\sec }^{-1}\frac{1}{\sqrt{1-x^2}}+{\cot }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)$

${\cos }^{-1}\sqrt{1-x^2}+{\tan }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$

${\sin }^{-1}x+{\sin }^{-1}x$

$2{\sin }^{-1}x$

${\sin }^{-1}2x\sqrt{1-x^2}$

$\therefore k=2x\sqrt{1-x^2}$