Question

if $si{n}^{-1}(x-1)+co{s}^{-1}(x-3)+ta{n}^{-1}\left(\frac{x}{2-x^2}\right)=co{s}^{-1}k+\pi$, then the value of $k$ is

• A. $1$
• B. $-\frac{1}{\sqrt{2}}$
• C. $\frac{1}{\sqrt{2}}$
• D. $noneofthese$

Answer 1

The correct option is C $\frac{1}{\sqrt{2}}$

Let $x=2$
Hence ${\sin }^{-1}\left(1\right)+{\cos }^{-1}(-1)+ta{n}^{-1}\left(\frac{2}{2-4}\right)$
$=\frac{\pi }{2}+\pi -\frac{\pi }{4}$
$=\frac{3\pi }{2}-\frac{\pi }{4}$
$=\frac{5\pi }{4}$
$={\cos }^{-1}\left(k\right)+\pi$
${\cos }^{-1}\left(k\right)=\frac{\pi }{4}$
$k=\cos \left(\frac{\pi }{4}\right)$
$=\frac{1}{\sqrt{2}}$