What is this trigonometric expression equal to?

Question

If $\frac{cos{x}}{cos{y}}=\frac{a}{b}$, then $(a\times tan{x}+b\times tan{y})$ equals
(A)$(a+b)cot{\frac{x+y}{2}}$
(B)$(a+b)tan\frac{x+y}{2}$
(C)$(a+b)(tan\frac{x}{2}+tan\frac{y}{2})$
(D)$(a+b)(cot\frac{x}{2}+cot\frac{y}{2})$
I believe know the trigonometric formulas but I am just not able to reduce the given expression to any of the options. Help.

Answer 1

If $x=y$ then the expression is $(a+b)\tan x$, so the only alternative that could possibly be true is (B): that is, $$\hbox{if}\quad\frac{\cos x}{\cos y}=\frac{a}{b}\quad\hbox{then}\quad a\tan x+b\tan y=(a+b)\tan\frac{x+y}{2}\ .$$ It remains to prove it.

We have $$\frac{a}{a+b}=\frac{\cos x}{\cos x+\cos y}\ ,\quad \frac{b}{a+b}=\frac{\cos y}{\cos x+\cos y}\ ,$$ so $${\rm LHS} =(a+b)\frac{\sin x+\sin y}{\cos x+\cos y} =(a+b)\frac{\sin\frac{x+y}{2}\cos\frac{x-y}{2}}{\cos\frac{x+y}{2}\cos\frac{x-y}{2}} ={\rm RHS}\ .$$