[SOLVED] Simple Identity Confirmation

**unstopabl3** · May 7th 2010, 12:18 AM

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

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

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

$=1-sin^2x-sin^2x$

$=1-2sin^2x$

Am I on the right track?

**Grandad** · May 7th 2010, 01:40 AM

Hello unstopabl3

Originally Posted by unstopabl3

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

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

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

$=1-sin^2x-sin^2x$

$=1-2sin^2x$

Am I on the right track?

Yes!

Grandad

**unstopabl3** · May 7th 2010, 02:05 AM

Great, thanks for fixing the Latex and your quick reply!

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