Question

What is ${\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{1}{3}\right)$ equal to?

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{6}$

Answer 1

Answer: (C)

$\frac{\pi }{4}$

We know the formula ${\tan }^{-1}a+{\tan }^{-1}b={\tan }^{-1}\left(\frac{a+b}{1-ab}\right)$

So ${\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{1}{3}\right)={\tan }^{-1}\left(\frac{\left(\frac{1}{2}\right)+\left(\frac{1}{3}\right)}{1-\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)}\right)={\tan }^{-1}\left(\frac{\left(\frac{5}{6}\right)}{\left(\frac{5}{6}\right)}\right)={\tan }^{-1}\left(1\right)=\frac{\pi }{4}$