Question

If $\tan A = \frac{1}{2}, \tan B = \frac{1}{3}$ then $\tan(A+2B)$ has the value

Answer 1

2

Answer 2

We are given:

$$\tan A = \frac{1}{2}, \quad \tan B = \frac{1}{3}$$

1. Find $\tan(A+B)$:

   $$\tan(A+B) = \frac{\tan A + \tan B}{1-\tan A\,\tan B} = \frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot\frac{1}{3}} = \frac{\frac{3+2}{6}}{1-\frac{1}{6}} = \frac{\frac{5}{6}}{\frac{5}{6}} = 1.$$

2. Find $\tan(A+2B)$ (using the addition formula again):

   $$\tan(A+2B) = \tan((A+B)+B) = \frac{\tan(A+B) + \tan B}{1-\tan(A+B)\,\tan B} = \frac{1 + \frac{1}{3}}{1- 1\cdot\frac{1}{3}} = \frac{\frac{4}{3}}{\frac{2}{3}} = 2.$$

Answer: $\tan(A+2B)=2$