Question

${\cos }^{-1}(x^2+\frac{1}{x^2}-1)+{\sin }^{-1}(x^2-\frac{1}{x^2})+{\tan }^{-1}\left(x^2\right)$ is equal to $(\text{where}x\in R-\{0\left\}\right)$

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B $\frac{\pi }{4}$

$x^2+\frac{1}{x^2}\ge 2$ but $x^2+\frac{1}{x^2}-1$ must be less than or equal to $1$

$\Rightarrow x^2=1$

So ${\cos }^{-1}\left(1\right)+{\sin }^{-1}\left(0\right)+{\tan }^{-1}\left(1\right)$

$=0+0+\frac{\pi }{4}=\frac{\pi }{4}$