A) 2
B) \[-1\]
C) 3
D) 1
Correct Answer: C
Solution :
[c] \[tan\alpha =\frac{1}{3}\] \[\And tan\frac{\beta }{2}=\frac{1}{2}\therefore \tan \beta =\frac{2\tan \frac{\beta }{2}}{1-{{\tan }^{2}}\frac{\beta }{2}}\] \[=\frac{2.\frac{1}{2}}{1.\frac{\beta }{2}}=\frac{1}{\frac{3}{4}}=\frac{4}{3}.\] Now, \[\tan \left( \alpha +\beta \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha .\tan \beta }\] \[=\frac{\frac{1}{3}+\frac{4}{3}}{1-\frac{1}{3}\times \frac{4}{3}}=\frac{\frac{5}{3}}{1-\frac{4}{9}}=\frac{\frac{5}{3}}{\frac{5}{9}}=\frac{5}{3}\times \frac{9}{5}=3\]
You need to login to perform this action.
You will be redirected in
3 sec
