Question

If ${\cot }^{-1}\left(\frac{1}{x}\right)+{\cos }^{-1}(-x)+{\tan }^{-1}x=\pi$ and ${\sin }^{-1}x<0,$ then the value of $\frac{(1-x^2{)}^{3/2}}{x^2}$ is

• A. $\frac{1}{2}$
• B. $1$
• C. $2$
• D. $\frac{1}{4}$

Answer 1

The correct option is C $2$

${\cot }^{-1}\left(\frac{1}{x}\right)+{\cos }^{-1}(-x)+{\tan }^{-1}x=\pi$

${\tan }^{-1}x+{\tan }^{-1}x+{\cos }^{-1}(-x)=\pi$

${\tan }^{-1}\frac{2x}{1-x^2}-{\tan }^{-1}\frac{\sqrt{1-x^2}}{x}=\pi$

${\tan }^{-1}\left[\frac{\frac{2x}{1-x^2}-\frac{\sqrt{1-x^2}}{x}}{1+\frac{2x}{1-x^2}\times \frac{\sqrt{1-x^2}}{x}}\right]=\pi$

$\Rightarrow \frac{\frac{2x^2-(1-x^2{)}^{\frac{3}{2}}}{x(1-x^2)}}{1+\frac{2}{\sqrt{1-x^2}}}=\tan \pi$

$\Rightarrow \frac{2x^2-(1-x^2{)}^{\frac{3}{2}}}{x(\sqrt{1-x^2}+2)}=0$

Hence, $x\sqrt{1-x^2}(\sqrt{1-x^2}+2)$ cannot be zero.

Hence, $2x^2=(1-x^2{)}^{\frac{3}{2}}$

$\Rightarrow \frac{(1-x^2{)}^{\frac{3}{2}}}{x^2}=2$