Question

Evaluate ${\tan }^{-1}\left(x\right)+co{t}^{-1}\left(x+1\right)=$

• A. ${\tan }^{-1}\left(x^2+1\right)$
• B. ${\tan }^{-1}\left(x^2+x\right)$
• C. ${\tan }^{-1}\left(x+1\right)$
• D. ${\tan }^{-1}\left(x^2+x+1\right)$

Answer 1

The correct option is D

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

Explanation for the correct option:

Find the value of the given Equation.

Consider the given Equation as

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

Rewrite the above Equation as

$y={\tan }^{-1}\left(x\right)+{\tan }^{-1}\left(\frac{1}{x+1}\right)\left[\because {\cot }^{-1}\left(x\right)={\tan }^{-1}\left(\frac{1}{x}\right)\right]$

Then,

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

Therefore,

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

Hence, option (D) is the correct answer.