Question

If tanα=1/2 and tan β=1/3,then find the value of α+β

Answer 1

$\tan\alpha = \frac{1}{2} \\ \\ \tan( \beta ) = \frac{1}{3} \\ \\ and \: \tan(\alpha + \beta) = \frac{ \tan\alpha + \tan\beta }{1 - \tan \alpha \tan\beta } \\ \\ \tan( \alpha + \beta ) = \frac{ \frac{1}{2} + \frac{1}{3} }{1 - \frac{1}{6} } \\ \\ = > \frac{ \frac{3 + 2}{6} }{ \frac{6 - 1}{6} } \\ \\ = > \frac{ \frac{5}{6} }{ \frac{5}{6} } \\ \\ = > 1 \\ \\ hence \: ( \alpha + \beta ) = { \tan }^{ - 1}(1 ) \\ \\ i.e \: \alpha + \beta = {45}degree$

Answer 2

Answer:

$→\tan( \alpha ) = \frac{1}{2}\:(given) \\ → \tan( \beta ) = \frac{1}{3}\:(given) \\→\boxed{ \tan( \alpha + \beta ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{1 - \tan( \alpha ) \tan( \beta ) }} ✓ \\→ \tan( \alpha + \beta ) = \frac{( \frac{1}{2} +\frac{1}{3})}{1 - ( \frac{1}{2} \times\frac{1}{3})} \\ → \tan( \alpha + \beta ) = \frac{ \frac{(3 + 2)}{6} }{1 - ( \frac{1}{6}) } \\ → \tan( \alpha + \beta ) = \frac{ \frac{5}{6} }{ \frac{5}{6} } \\ → \tan( \alpha + \beta ) = 1 \\→ ( \alpha + \beta ) = \tan ^{ - 1} (1) \\ → \boxed{( \alpha + \beta ) = \frac{\pi}{4} }✓$

• π/4 is the right answer.