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Q. If tan A = 1/2, tan B = 1/3; then what is the value of tan (2A + B) ?

Trigonometric Functions

Solution:

Given values are $\tan \, A = \frac{1}{2} \tan B = \frac{1}{3} \tan\left(2A+B\right) = \frac{\tan2A +\tan B}{1-\tan2A \tan B}$
and, $\tan2A = \frac{2 \tan A}{1- \tan^{2}A} = \frac{2 \frac{1}{2}}{1- \frac{1}{4}} = \frac{4}{3} $
So, $\tan\left(2A + B \right) = \frac{\frac{4}{3} + \frac{1}{3} }{1- \frac{4}{3} \times\frac{1}{3}} = \frac{ \frac{5}{3}}{\frac{5}{9}}=3 $