Question

The value of ${\tan }^{-1}x+{\cot }^{-1}x$ is.

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

Answer 1

The correct option is C $\frac{\pi }{2}$

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

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

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

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

Since the term inside the bracket tends to $\infty$, the answer is $\frac{\pi }{2}$