Question

If $\tan \left(co{s}^{-1}x\right)=\sin \left({\sec }^{-1}\right(\sqrt{5}\left)\right)$, then $x=$

• A. $\sqrt{\frac{5}{3}}$
• B. $\frac{\sqrt{5}}{3}$
• C. $\sqrt{\frac{5}{2}}$
• D. $\frac{\sqrt{5}}{2}$

Answer 1

The correct option is B $\frac{\sqrt{5}}{3}$

If $tan\left(co{s}^{-1}x\right)=sin\left(se{c}^{-1}\sqrt{5}\right)$
$\therefore Let\alpha =co{s}^{-1}x$
$cos\alpha =x$
$\therefore tan\alpha =\frac{\sqrt{1-x^2}}{x}$
$\therefore tan\left(co{s}^{-1}x\right)=\frac{\sqrt{1-x^2}}{x}=\frac{2}{\sqrt{5}}$
$\Rightarrow 5(1-x^2)=4x^2$
$5=9x^2$
$x=\pm \frac{\sqrt{5}}{3}$