Simplify trig equation

**zapparage** · August 26th 2008, 11:05 AM

How do I simply $(1+tan^2A)(1-sin^2A)$

**flyingsquirrel** · August 26th 2008, 11:12 AM

Hello,

Originally Posted by zapparage

How do I simply $(1+tan^2A)(1-sin^2A)$

You can start by using $1+\tan^2A=\frac{1}{\cos^2A}$ . If you're not supposed to know this identity you can start as follows :

$<br /> (1+\tan^2A)(1-\sin^2A)=(1+\tan^2A)\cos^2A=\left(1+\frac{\sin^2A} {\cos^2A}\right)\cos^2A=\ldots$

**11rdc11** · August 26th 2008, 11:13 AM

It Equals 1, it may take me a lil while to post the answer since I'm trying to learn latex lol.

**11rdc11** · August 26th 2008, 11:17 AM

${1-sin^2A} = cos^2A$

**11rdc11** · August 26th 2008, 11:41 AM

Here is another way you could do it,

$1- sin^2A +tan^2A-tan^2Asin^2A= 1 -sin^2A +\frac{sin^2A}{cos^2A} -\frac{sin^4A}{cos^2A}$

$\frac{cos^2A-sin^2Acos^2A+sin^2A-sin^4A}{cos^2A}$

$\frac{cos^2A(1-sin^2A)+sin^2A(1-sin^2A)}{cos^2A}$

$\frac{cos^4A+sin^2Acos^2A}{cos^2A}$

$\frac {cos^2A(cos^2A + sin^2A)}{cos^2A}$

$cos^2A + sin^2A= 1$

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