Browse Questions

# If $\tan \theta=\large\frac{1}{2}$ and $\tan\phi=\large\frac{1}{3}$,then the value of $\theta+\phi$ is

$\begin{array}{1 1}(A)\;\large\frac{\pi}{6}&(B)\;\pi\\(C)\;0&(D)\;\large\frac{\pi}{4}\end{array}$

Toolbox:
• $\tan(A+B)=\large\frac{\tan A+\tan B}{1-\tan A.\tan B}$
Given :
$\tan \theta=\large\frac{1}{2}$ and $\tan\phi=\large\frac{1}{3}$
$\therefore \tan(\theta+\phi)=\large\frac{\tan \theta+\tan\phi}{1-\tan \theta.\tan \phi}$
$\Rightarrow \large\frac{\Large\frac{1}{2}+\frac{1}{3}}{1-\Large\frac{1}{2}\times \frac{1}{3}}$
$\Rightarrow \large\frac{\Large\frac{5}{6}}{\Large\frac{5}{6}}$
$\Rightarrow \tan (\theta+\phi)=1$
$\therefore \theta+\phi=\large\frac{\pi}{4}$
Hence (D) is the correct answer.