## Trig Identity prob: show that cot^2x sinx/(1-sinx)(1+sinx) = 1/sinx

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.
jperkins0236
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### Trig Identity prob: show that cot^2x sinx/(1-sinx)(1+sinx) = 1/sinx

Hi everyone, Im struggling with a trig identity problem, and I just can't seem to figure out how to solve it.

The problem says show that

cot^2x sinx/(1-sinx)(1+sinx) = 1/sinx

The furthest I could get was cot^2x sinx/cos^x and I'm not sure how to further complete this problem to prove the equivalency.

Any help would be greatly appreciated... Thank you!

little_dragon
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Joined: Mon Dec 08, 2008 5:18 pm
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### Re: Trig Identity prob: show that cot^2x sinx/(1-sinx)(1+sinx) = 1/sinx

show that cot^2x sinx/(1-sinx)(1+sinx) = 1/sinx
i'm taking your eqn to mean this:

. . . . .$\dfrac{\cot^2(x)\, \sin(x)}{(1\, -\, \sin(x))\, (1\, +\, \sin(x))}\, =\, \dfrac{1}{\sin(x)}$

use the way they show here: http://www.purplemath.com/modules/proving.htm
the LHS is harder so start there & convert everything to sines and cosines:

. . . . .$\left(\dfrac{\cos^2(x)}{\sin^2(x)}\right)\, \left(\dfrac{\sin(x)}{1\, -\, \sin^2(x)} \right)\, =\, \dfrac{1}{\sin(x)}$

then apply the identity to the second denominator
then cancel