Question

If $\cot \left({\sin }^{-1}\sqrt{\frac{13}{17}}\right)=\sin \left({\tan }^{-1}x\right)$, then $x$ is equal to

• A. $\frac{2}{\sqrt{13}}$
• B. $\frac{-2}{3}$
• C. $\frac{-2}{\sqrt{13}}$
• D. $\frac{2}{3}$

Answer 1

Answer: (D)

Given that, $\cot \left({\sin }^{-1}\sqrt{\frac{13}{17}}\right)=\sin \left({\tan }^{-1}x\right)$ Note that $x$ must be positive.
Put ${\sin }^{-1}\sqrt{\frac{13}{17}}=\theta$.

$\therefore \text{ L.H.S. }=\frac{2}{\sqrt{13}}$
Put ${\tan }^{-1}x=\varphi$
$\therefore \text{ R.H.S. }=\frac{x}{\sqrt{1+x^2}}$
So, given equation is $\frac{2}{\sqrt{13}}=\frac{x}{\sqrt{1+x^2}}$ (on squaring)
$\Rightarrow x=\frac{2}{3}(\text{ As }x>0)$