Question

Evaluate ${\tan }^{-1}\left(\frac{1-x^2}{2x}\right)+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=$

• A. $\frac{\mathrm{\pi }}{4}$
• B. $\frac{\mathrm{\pi }}{2}$
• C. $\mathrm{\pi }$
• D. $0$

Answer 1

The correct option is B

$\frac{\mathrm{\pi }}{2}$

Explanation for the correct option:

Find the value of the given Equation.

Consider the given Equation as

$y={\tan }^{-1}\left(\frac{1-x^2}{2x}\right)+{\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\to \left(i\right)$

We know that,

$$\begin{gathered} {\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2{\tan }^{-1}\left(x\right) \\ ={\tan }^{-1}\left(\frac{2x}{1-x^2}\right) \end{gathered}$$

Then, the equation $\left(i\right)$ becomes

$y={\tan }^{-1}\left(\frac{1-x^2}{2x}\right)+{\tan }^{-1}\left(\frac{2x}{1-x^2}\right)$

Then,

$$\begin{gathered} y={\tan }^{-1}\left(\frac{\frac{1-x^2}{2x}+\frac{2x}{1-x^2}}{1-\left(\frac{1-x^2}{2x}\right)\left(\frac{2x}{1-x^2}\right)}\right)\left[\because {\tan }^{-1}A+{\tan }^{-1}B={\tan }^{-1}\left(\frac{A+B}{1-\mathrm{AB}}\right)\right] \\ y={\tan }^{-1}\left[\frac{\frac{1-x^2}{2x}+\frac{2x}{1-x^2}}{1-1}\right] \\ y={\tan }^{-1}\left[\frac{\frac{1-x^2}{2x}+\frac{2x}{1-x^2}}{0}\right] \\ y={\tan }^{-1}\left(\infty \right) \\ y=\frac{\mathrm{\pi }}{2} \end{gathered}$$

Hence, option (B) is the correct answer.