# Thread: Prove this trig identity.

1. ## Prove this trig identity.

Can someone solve it step by step. I tried two ways to solve it but didn't work.

2. 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?

3. $\displaystyle \frac{\sec^2{x}}{2 - \sec^2{x}} =$

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

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

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

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

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

$\displaystyle \sec(2x)$

4. Hello, mwok!

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

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

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

5. Thanks!

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

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

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

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

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

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

$\displaystyle \sec(2x)$
Why do you multiply with the cos/sin?

7. Originally Posted by mwok
Why do you multiply with the cos/sin?
I didn't ... I multiplied by $\displaystyle \frac{\cos^2{x}}{\cos^2{x}}$