Question

${\tan }^{-1}a+{\tan }^{-1}b,$ where $a>0,b>0,ab>1,$ is equal to

• A. ${\tan }^{-1}\frac{a+b}{1-ab}$
• B. ${\tan }^{-1}\frac{a+b}{1-ab}-\pi$
• C. $\pi +{\tan }^{-1}\frac{a+b}{1-ab}$
• D. none of these

Answer 1

Answer: (C)

$\pi +{\tan }^{-1}\frac{a+b}{1-ab}$

Given, ${\tan }^{-1}\left(a\right)+{\tan }^{-1}\left(b\right)$
$={\tan }^{-1}\left(\frac{a+b}{1-ab}\right)$
However, $ab>1$
Therefore,
$\pi +{\tan }^{-1}\left(\frac{a+b}{1-ab}\right)$