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.
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user140087
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1a good way when given options to choose from is so test the function at where it is undefined. Here $x,y=\pi/2,\pi/2$. That immediately removes the third and fourth options. By checking $x=y=0$, you can remove the first option as well. Now that you know it must be the second, work towards reducing it to that form. – Guy Apr 03 '14 at 11:35
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http://math.stackexchange.com/questions/510108/if-frac-cos-x-cos-y-fracab-then-a-tan-x-b-tan-y-equals – lab bhattacharjee Apr 04 '14 at 03:51
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@labbhattacharjee I wasn't aware that this question has already been asked. What can I do now? – user140087 Apr 05 '14 at 11:00
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@user140087, I don't think you've something to do :). – lab bhattacharjee Apr 05 '14 at 12:55
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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 $${\rm LHS} =(a+b)\frac{\sin x+\sin y}{\cos x+\cos y} =(a+b)\frac{\sin\frac{x+y}{2}\cos\frac{x-y}{2}}{\cos\frac{x+y}{2}\cos\frac{x-y}{2}} ={\rm RHS}\ .$$
David
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