Question
If $$\alpha + \beta = \frac{\pi }{2}\,{\text{and }}\beta + \gamma = \alpha ,$$ then $$\tan \alpha $$ equals
A.
$$2\left( {\tan \beta + \tan \gamma } \right)$$
B.
$${\tan \beta + \tan \gamma }$$
C.
$${\tan \beta + 2\tan \gamma }$$
D.
$${2\tan \beta + \tan \gamma }$$
Answer :
$${\tan \beta + 2\tan \gamma }$$
Solution :
Given that $$\alpha + \beta = \frac{\pi }{2}\,$$
$$\eqalign{
& \Rightarrow \,\alpha = \frac{\pi }{2} - \beta \cr
& \Rightarrow \,\,\tan \alpha = \tan \left( {\frac{\pi }{2} - \beta } \right) = \cot \beta = \frac{1}{{\tan \beta }} \cr
& \Rightarrow \,\,\tan \alpha \tan \beta = 1 \cr
& \Rightarrow \,\,1 + \tan \alpha \tan \beta = 2. \cr
& \,\therefore \,\,\tan \left( {\alpha - \beta } \right) = \frac{{\tan \alpha - \tan \beta }}{{1 + \tan \alpha \tan \beta }} \cr
& \Rightarrow \,\,\tan \gamma = \frac{{\tan \alpha - \tan \beta }}{2} \cr
& \Rightarrow \,\,2\tan \gamma = \tan \alpha - \tan \beta \cr
& \Rightarrow \,\,\tan \alpha = 2\tan \gamma + \tan \beta \cr} $$