<s>[INST] 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.
 [/INST] 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.
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