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

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- Aug 26th 2008, 11:05 AMzapparageSimplify trig equation
How do I simply $\displaystyle (1+tan^2A)(1-sin^2A)$

- Aug 26th 2008, 11:12 AMflyingsquirrel
- Aug 26th 2008, 11:13 AM11rdc11
It Equals 1, it may take me a lil while to post the answer since I'm trying to learn latex lol.

- Aug 26th 2008, 11:17 AM11rdc11
$\displaystyle {1-sin^2A} = cos^2A$

- Aug 26th 2008, 11:41 AM11rdc11
Here is another way you could do it,

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

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

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

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

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

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