# Simplify trig equation

• Aug 26th 2008, 11:05 AM
zapparage
Simplify trig equation
How do I simply $\displaystyle (1+tan^2A)(1-sin^2A)$
• Aug 26th 2008, 11:12 AM
flyingsquirrel
Hello,
Quote:

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

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

$\displaystyle (1+\tan^2A)(1-\sin^2A)=(1+\tan^2A)\cos^2A=\left(1+\frac{\sin^2A} {\cos^2A}\right)\cos^2A=\ldots$
• Aug 26th 2008, 11:13 AM
11rdc11
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 AM
11rdc11
$\displaystyle {1-sin^2A} = cos^2A$
• Aug 26th 2008, 11:41 AM
11rdc11
Here is another way you could do it,

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

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

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

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

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

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