Identities Problem

**Hellooo** · January 23rd 2010, 07:25 AM

How someone help me prove the identity below?

(1-tan^2X)(1-cos^2X)=[(sin^2X-4sin^4X)/(1-sin^2X)]

Thanks in advance! =)

**Grandad** · January 23rd 2010, 08:41 AM

Hello Hellooo

Welcome to Math Help Forum!

Originally Posted by Hellooo

How someone help me prove the identity below?

(1-tan^2X)(1-cos^2X)=[(sin^2X-4sin^4X)/(1-sin^2X)]

Thanks in advance! =)

I think the 4 on the RHS should be a 2. Here's my answer:

$(1-\tan^2x)(1-\cos^2x)$

$=\left(1-\frac{\sin^2x}{\cos^2x}\right)\sin^2x$ , using $\tan x = \frac{\sin x}{\cos x}$ and $1 - \cos^2x = \sin^2 x$

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

$=\frac{(1-2\sin^2x)\sin^2x}{1-\sin^2x}$ , using $\cos^2x =1-\sin^2x$

$=\frac{\sin^2x-2\sin^4x}{1-\sin^2x}$

Grandad

**Hellooo** · January 23rd 2010, 04:24 PM

Oh thanks,
I was thinking..why was there a 4 there..

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