If $\frac{\cos x}{\cos y}=\frac{a}{b}$ then $a\tan x +b\tan y$ equals If $$\frac{\cos x}{\cos y}=\frac{a}{b}$$ Then $$a \cdot\tan x +b \cdot\tan y$$  Equals to (options below): 
(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)  $(a+b)(\cot\frac{x}{2}+\cot\frac{y}{2})$ 
My approach : 
$$\frac{\cos x}{\cos y} = \frac{a}{b} $$
[ Using componendo and dividendo ] 
$$\frac{\cos x +\cos y}{\cos x -\cos y} = \frac{a+b}{a-b}$$
$$=\frac{2\cos\frac{x+y}{2}\cos\frac{x-y}{2}}{2\sin\frac{x+y}{2}\sin\frac{y-x}{2}}$$
I'm stuck, I'd aprecciate any suggestions. Thanks.
 A: I am hoping that my calculations haven't gone wrong. Here is a method to proceed.
From your method you have $$\frac{2\cos\frac{x+y}{2}\cos\frac{x-y}{2}}{2\sin\frac{x+y}{2}\sin\frac{y-x}{2}} = -\cot\Bigl(\frac{x+y}{2}\Bigr)\cdot\Bigl(\frac{x-y}{2}\Bigr)=\frac{a+b}{a-b}$$
Now note that 
\begin{align*}
\tan(x) &=\tan\left(\frac{x+y}{2} + \frac{x-y}{2}\right)
\\ &=\frac{\tan\left(\frac{x+y}{2}\right)+\tan\left(\frac{x-y}{2}\right)}{1-\tan\left(\frac{x+y}{2}\right)\cdot\tan\left(\frac{x-y}{2}\right)} \\ &=\frac{\tan\left(\frac{x+y}{2}\right)+\tan\left(\frac{x-y}{2}\right)}{1-\frac{a-b}{a+b}} \\ &=\frac{a+b}{2b} \times \tan\left(\frac{x+y}{2}\right)+\tan\left(\frac{x-y}{2}\right)
\end{align*}
Similary $$\tan(y) =\frac{a+b}{2a} \times \biggl\{\tan\left(\frac{x+y}{2}\right)-\tan\left(\frac{x-y}{2}\right)\biggr\}$$
Now just multiply the $\tan(x)$ quantity by $a$ and $\tan(y)$ quantity by $b$ and add both the sides and see if you get the answer.
A: So, we have $$\frac a{\cos x}=\frac b{\cos y}=\frac{a+b}{\cos x+\cos y}$$
$$\implies a\tan x+b\tan y=\frac a{\cos x}\cdot\sin x+\frac b{\cos y}\cdot\sin y$$
Putting the values of $\displaystyle\frac a{\cos x},\frac b{\cos y}$
$$a\tan x+b\tan y=(\sin x+\sin y)\frac{(a+b)}{\cos x+\cos y}$$
Now, $\displaystyle \cos x+\cos y=2\cos\frac{x+y}2\cos\frac{x-y}2$ and 
$\displaystyle \sin x+\sin y=2\sin\frac{x+y}2\cos\frac{x-y}2$
