How do you prove the identity $\frac{{tan}^{2} x}{sec x + 1} = \frac{1 - cos x}{cos x}$ $tan^2x/(secx+1)= (1-cosx)/cosx$ ?

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How do you prove the identity $\frac{{tan}^{2} x}{sec x + 1} = \frac{1 - cos x}{cos x}$ $tan^2x/(secx+1)= (1-cosx)/cosx$ ?

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Answer 1 · 2015-02-07T13:29:49

You can change:
$tan (x)$ $tan(x)$ into $\frac{sin (x)}{cos (x)}$ $sin(x)/cos(x)$ and
$sec (x)$ $sec(x)$ into $\frac{1}{cos (x)}$ $1/cos(x)$
Giving:
enter image source here

hope it helps

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How do you prove the identity $\frac{{tan}^{2} x}{sec x + 1} = \frac{1 - cos x}{cos x}$ $tan^2x/(secx+1)= (1-cosx)/cosx$ ?

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How do you prove the identity tan^2x/(secx+1)= (1-cosx)/cosxtan2xsecx+1=1−cosxcosxtan^2x/(secx+1)= (1-cosx)/cosx?

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How do you prove the identity $\frac{{tan}^{2} x}{sec x + 1} = \frac{1 - cos x}{cos x}$ $tan^2x/(secx+1)= (1-cosx)/cosx$ ?