Basic Identities

**anandraj** · July 14th 2010, 08:31 PM

Hello there!

I have this problem on proving identities. I've been working on it for hours, but can't seem to get it right. The sad thing is that it's not even a double angle formula

cos x + sin x = [cos x / (1 - tan x)] + [sin x / (1 - cot x)]

Any help would be greatly appreciated. Thanks!

**red_dog** · July 14th 2010, 10:26 PM

Start from the right side and write $\tan x$ and $\cot x$ in terms of $\sin x$ and $\cos x$ .

**anandraj** · July 14th 2010, 10:42 PM

Originally Posted by red_dog

Start from the right side and write $\tan x$ and $\cot x$ in terms of $\sin x$ and $\cos x$ .

Hi red_dog,

Yes I did that but I end up getting an extremely long sequence which ends up with a factor of three for sin and cos, that's when I get stuck.

**red_dog** · July 14th 2010, 10:49 PM

$\displaystyle\frac{\cos x}{1-\tan x}+\frac{\sin x}{1-\cot x}=$

$=\displaystyle\frac{\cos x}{1-\displaystyle\frac{\sin x}{\cos x}}+\frac{\sin x}{1-\displaystyle\frac{\cos x}{\sin x}}=$

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

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

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

$=\displaystyle\frac{(\cos x-\sin x)(\cos x+\sin x)}{\cos x-\sin x}=\cos x+\sin x$

**anandraj** · July 14th 2010, 11:00 PM

Hey thanks red_dog!

I realised my mistake now. I cross multiplied the fourth line (which was why i was getting the cube) instead of multiplying -1 to the second term.

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