# Prove this trig identity.

• Nov 26th 2008, 06:57 AM
mwok
Prove this trig identity.
Can someone solve it step by step. I tried two ways to solve it but didn't work.
• Nov 26th 2008, 07:04 AM
Smancer
Its not clear to me what your question is, what are you supposed to show?

It is clear that your steps are not correct. Can you state the question exactly how it is given to you?
• Nov 26th 2008, 07:10 AM
skeeter
$\frac{\sec^2{x}}{2 - \sec^2{x}} =$

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

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

multiply by $\frac{\cos^2{x}}{\cos^2{x}}$ ...

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

$\frac{1}{\cos(2x)} =$

$\sec(2x)$
• Nov 26th 2008, 07:15 AM
Soroban
Hello, mwok!

Is this it? . $\sec2\theta \:=\:\frac{\sec^2\!\theta}{2-\sec^2\!\theta}$

The left side is: . $\sec2\theta \;=\;\frac{1}{\cos2\theta} \;=\;\frac{1}{2\cos^2\!\theta - 1}$

Divide top and bottom by $\cos^2\!\theta\!:\quad \frac{\dfrac{1}{\cos^2\!\theta}}{\dfrac{2\cos^2\!\ theta}{\cos^2\!\theta} - \dfrac{1}{\cos^2\!\theta}} \;=\;\frac{\sec^2\!\theta}{2 - \sec^2\!\theta}$

• Nov 26th 2008, 08:15 AM
mwok
Thanks!
• Nov 26th 2008, 09:10 AM
mwok
Quote:

Originally Posted by skeeter
$\frac{\sec^2{x}}{2 - \sec^2{x}} =$

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

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

multiply by $\frac{\cos^2{x}}{\cos^2{x}}$ ...

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

$\frac{1}{\cos(2x)} =$

$\sec(2x)$

Why do you multiply with the cos/sin?
• Nov 26th 2008, 09:38 AM
skeeter
Quote:

Originally Posted by mwok
Why do you multiply with the cos/sin?

I didn't ... I multiplied by $\frac{\cos^2{x}}{\cos^2{x}}$