Question

If $\tan \theta =\frac{1}{2},\tan \varphi =\frac{1}{3}$ then the value of $\tan (\theta +\varphi )$ is,

• A. $\frac{\pi }{6}$
• B. $\pi$
• C. 0
• D. $\frac{\pi }{4}$

Answer 1

The correct option is D $\frac{\pi }{4}$

we have,

$\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x\tan y}$

Given,

$\tan (\theta +\varphi )=\frac{\tan \theta +\tan \varphi }{1-\tan \theta \tan \varphi }$

$=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\times \frac{1}{3}}$

$=\frac{\frac{5}{6}}{\frac{5}{6}}$

$=1$

$\Rightarrow \tan (\theta +\varphi )=1$

$\therefore \theta +\varphi ={\tan }^{-1}1=\frac{\pi }{4}$