Can someone solve it step by step. I tried two ways to solve it but didn't work.
$\displaystyle \frac{\sec^2{x}}{2 - \sec^2{x}} =$
$\displaystyle \frac{1 + \tan^2{x}}{2 - (1 + \tan^2{x})} =$
$\displaystyle \frac{1 + \tan^2{x}}{1 - \tan^2{x}}$
multiply by $\displaystyle \frac{\cos^2{x}}{\cos^2{x}}$ ...
$\displaystyle \frac{\cos^2{x} + \sin^2{x}}{\cos^2{x} - \sin^2{x}} =$
$\displaystyle \frac{1}{\cos(2x)} =$
$\displaystyle \sec(2x)$
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 $\displaystyle \cos^2\!\theta\!:\quad \frac{\dfrac{1}{\cos^2\!\theta}}{\dfrac{2\cos^2\!\ theta}{\cos^2\!\theta} - \dfrac{1}{\cos^2\!\theta}} \;=\;\frac{\sec^2\!\theta}{2 - \sec^2\!\theta}$