Question:
If $\tan \theta=\frac{1}{2}$ and $\tan \phi=\frac{1}{3}$, then the value of $\theta+\phi$ is
(a) $\frac{\pi}{6}$
(b) $\pi$
(c) 0
(d) $\frac{\pi}{4}$
Solution:
It is given that $\tan \theta=\frac{1}{2}$ and $\tan \phi=\frac{1}{3}$.
Now,
$\tan (\theta+\phi)=\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}$
$=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}$
$=\frac{\frac{5}{6}}{\frac{5}{6}}$
= 1
$\therefore \theta+\phi=\frac{\pi}{4} \quad\left(\tan \frac{\pi}{4}=1\right)$
Hence, the correct answer is option D.