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

Printable View

- Nov 26th 2008, 06:57 AMmwokProve 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 AMSmancer
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 AMskeeter
$\displaystyle \frac{\sec^2{x}}{2 - \sec^2{x}} =$

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

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

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

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

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

$\displaystyle \sec(2x)$ - Nov 26th 2008, 07:15 AMSoroban
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 $\displaystyle \cos^2\!\theta\!:\quad \frac{\dfrac{1}{\cos^2\!\theta}}{\dfrac{2\cos^2\!\ theta}{\cos^2\!\theta} - \dfrac{1}{\cos^2\!\theta}} \;=\;\frac{\sec^2\!\theta}{2 - \sec^2\!\theta}$

- Nov 26th 2008, 08:15 AMmwok
Thanks!

- Nov 26th 2008, 09:10 AMmwok
- Nov 26th 2008, 09:38 AMskeeter