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B18JDM · Oct 17, 2010, 06:03 PM

tan(x-y)=(cos(y)-sin(y)cot(x))/(cot(x)cos(y)+sin(y)) I've tried numerous ways and am missing something. Please help.

lemon14 · Oct 18, 2010, 05:43 AM

there could be used 2 methods:

[math]\tan(x-y)=\frac{\cosy - \sin y\cot x}{\cot x \cos y + \sin y}[math]

the first one:
$\tan(x-y)=\frac{\tan x- \tan y}{1+ \tan x\tan y}=\frac{\frac{\sin x}{\cos x}-\frac{\sin y}{\cos y}}{1+ \frac{\sin x}{\cos x}\frac{\sin y}{\cos y}}=\frac{{\sin x\cos y - \sin y\cos x}{\cos x\cos y}}{{\cos x\cos y+\sin x\sin y}{\cos x\cos y}}=\frac{sin(x-y)}{cos(x-y)}=\tan(x-y)$

the second one:
$\frac{\cos y -\sin y \frac{\cos x}{\sin x}}{\frac{\cos x}{\sin x}+\sin y}=\frac{\cos y\sin x-\sin y\cos x}{\cos x\cos y + \sin x\sin y}$
and from this point I think you got it

Good luck! :)

lemon14 · Oct 18, 2010, 05:46 AM

there is
$\frac{\frac{\sin x\cos y - \siny\cos x}{\cos x\cos y}}{\frac{\cos x\cos y+\sin x\sin y}{\cos x\cos y}}$

sorry :(