Question

The value of x, where x > 0 and $tan\left(se{c}^{-1}\left(\frac{1}{x}\right)\right)=sin\left(ta{n}^{-1}2\right)$ is

• A. √5
  
• B. √5/3
• C. 1
• D. 2/3

Answer 1

The correct option is B

√5/3

$tan\left(se{c}^{-1}\left(\frac{1}{x}\right)\right)=sin\left(ta{n}^{-1}2\right)$
$\Rightarrow tan\left(ta{n}^{-1}\frac{\sqrt{1-x^2}}{x}\right)=sin\left(si{n}^{-1}\frac{2}{\sqrt{1+2^2}}\right)$
$(\because ta{n}^{-1}x=si{n}^{-1}\frac{x}{\sqrt{1-x^2}})$
$\Rightarrow \frac{\sqrt{1-x^2}}{x}=\frac{2}{\sqrt{5}}$
$\Rightarrow \sqrt{5}\sqrt{1-x^2}=2x$
$\Rightarrow 4x^2=5(1-x^2)$
$\Rightarrow x^2=\frac{5}{9}$
$\Rightarrow x=\frac{\sqrt{5}}{3}$ $(\because x>0given)$