Question

If ${\theta }_1$ and ${\theta }_2$ are acute angles, such that $\tan {\theta }_1=\frac{1}{2},\tan {\theta }_2=\frac{1}{3}$ and $\tan \left({\theta }_1+{\theta }_2\right)=\frac{\tan {\theta }_1+\tan {\theta }_2}{1-\tan {\theta }_1\tan {\theta }_2}$, then find $\left({\theta }_1+{\theta }_2\right)$.

Answer 1

Evaluating the given expression:

Using the trigonometric identity

$\tan \left({\theta }_1+{\theta }_2\right)=\frac{\tan {\theta }_1+\tan {\theta }_2}{1-\tan {\theta }_1\tan {\theta }_2}$

$\tan \left({\theta }_1+{\theta }_2\right)=\frac{{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}}{1-{\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{3}}}=\frac{{\displaystyle \frac{3+2}{6}}}{1-{\displaystyle \frac{1}{6}}}$

$\tan \left({\theta }_1+{\theta }_2\right)=\frac{{\displaystyle \frac{5}{6}}}{{\displaystyle \frac{5}{6}}}=1$

We know that,

$\tan 45°=1$

Thus, $\left({\theta }_1+{\theta }_2\right)=45°$

Hence, the value of the given expression is $45°$.