Question

If ${\cos }^{−1}x=Î\pm ,(0<x<1){\sin }^{−1}\left(2x\sqrt{1−x^2}\right)+{\sec }^{−1}\left(\frac{1}{2x^2−1}\right)=\frac{2Ï€}{3}$, then ${\tan }^{−1}(2x)$ equals:

• A. $\frac{Ï€}{2}$
• B. $\frac{Ï€}{3}$
• C. $\frac{Ï€}{4}$
• D. $\frac{Ï€}{6}$

Answer 1

Answer: (B)

Given that ${\cos }^{−1}x=Î\pm ,0<x<1…(i)$
$⇒x=\cos Î\pm$ $∴{\sin }^{−1}\left(2x\sqrt{1−x^2}\right)+{\sec }^{−1}\left(\frac{1}{2x^2−1}\right)=\frac{2Ï€}{3}$
$⇒{\sin }^{−1}\left(2\cos Î\pm \sqrt{1−{\cos }^2Î\pm }\right)+{\sec }^{−1}\left(\frac{1}{2{\cos }^2Î\pm −1}\right)=\frac{2Ï€}{3}$
$⇒{\sin }^{−1}\left(\sin 2Î\pm \right)+{\sec }^{−1}\left(\sec 2Î\pm \right)=\frac{2Ï€}{3}$
$⇒2Î\pm +2Î\pm =\frac{2Ï€}{3}$
$⇒Î\pm =\frac{Ï€}{6}$ From (i)
Now , $x=\cos \frac{Ï€}{6}=\frac{\sqrt{3}}{2}$
$⇒2x=\sqrt{3}$
$∴{\tan }^{−1}\left(2x\right)={\tan }^{−1}\left(\sqrt{3}\right)$
$=\frac{Ï€}{3}$