# Trig Identities

• January 24th 2009, 01:37 PM
casey_k
Trig Identities
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• January 24th 2009, 01:44 PM
Jhevon
Quote:

Originally Posted by casey_k
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apply these identities to the left hand side:

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

then apply these to the right:

$\tan x = \frac {\sin x}{\cos x}$
$1 - \cos^2 x = \sin^2 x$
• January 24th 2009, 01:45 PM
Jester
Quote:

Originally Posted by casey_k
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$\frac{ \cot x}{1 - \sin ^2 x} { ? = } \frac{ \tan x}{1 - \cos ^2 x}$

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

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

where $? =$ is a question - is this true. Do you see it from here?