Question

If $co{t}^{-1}x+si{n}^{-1}\left(\frac{1}{\sqrt{5}}\right)=\frac{\pi }{4}$, then $x$ is equal to

• A. $3$
• B. $\frac{1}{\sqrt{5}}$
• C. $0$
• D. None of these

Answer 1

Answer: (A)

$\because co{t}^{-1}x+si{n}^{-1}\frac{1}{\sqrt{5}}=\frac{\pi }{4}$
$\Rightarrow ta{n}^{-1}\frac{1}{x}+ta{n}^{-1}\frac{1}{2}=ta{n}^{-1}1$
$\Rightarrow ta{n}^{-1}\frac{1}{x}=ta{n}^{-1}1-ta{n}^{-1}\frac{1}{2}=ta{n}^{-1}\left(\frac{1-\frac{1}{2}}{1+\frac{1}{2}}\right)$
$\Rightarrow ta{n}^{-1}\frac{1}{x}=ta{n}^{-1}\frac{1}{3}$
$\Rightarrow x=3$