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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.

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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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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 $\displaystyle \sin x+\sin y=2\sin\frac{x+y}2\cos\frac{x-y}2$

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  • $\begingroup$ @sultan, can you take it from here? $\endgroup$ – lab bhattacharjee Sep 30 '13 at 18:52

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