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bellatrix · Nov 25, 2010, 02:50 AM

Verifying/proving of the trigonometric identities

lemon14 · Nov 25, 2010, 07:02 AM

I looked upon it, but I think that it's something wrong with the exercise.

$\frac{\cos x}{\sec x} - \frac{\cos x}{\tan ^2 x}=\cot ^2 x$

$\sec x= \frac{1}{\cos x}$

$\frac{\cos x}{{1}{\cos x}} - \frac{\cos x}{{\sin ^2 x}{\cos ^2 x}}=$

$\cos ^2 x - \frac{\cos ^3 x}{\sin ^2 x}=$

$\frac{\cos ^2 x \sin ^2 x - \cos ^3 x}{\sin ^2 x}=$

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

$\frac{\cos ^2 x}{\sin ^2 x}= \cot ^2 x$

$\cot ^2 x (\sin ^2 - \cos x)$

if there had been $\sin ^ 2 x + \cos ^2 x$ instead of $\sin ^2 x - \cos x$ it would have been true, but maybe I did some mistakes. On the whole, the idea is to use the formulas and calculate.

galactus · Nov 25, 2010, 07:57 AM

This is not a valid identity.

bellatrix · Nov 25, 2010, 09:21 PM

I also came to that answer and didn't get a valid identity, I think there is something wrong in the given, thanks anyway!