Question

If $\sin 2x=\alpha -1$ and $\cos 2x=\beta -1$, then the value of $\frac{{\sec }^{2}x\left[\right({\cos }^{2}x-{\sin }^{2}x)-2\sin x\cos x]}{1+\sin 2x}$ is equal to

• A. $2(\alpha -\beta )$
• B. $\frac{2}{\alpha }-\frac{2}{\beta }$
• C. $\frac{2}{\beta }-\frac{2}{\alpha }$
• D. $2(\alpha +\beta )$

Answer 1

The correct option is B $\frac{2}{\alpha }-\frac{2}{\beta }$

Given $\sin 2x=\alpha -1$
and $\cos 2x=\beta -1$
Consider, $\frac{{\sec }^{2}x\left[\right({\cos }^{2}x-{\sin }^{2}x)-2\sin x\cos x]}{1+\sin 2x}$
$=\frac{2(\cos 2x-\sin 2x)}{2{\cos }^{2}x(1+\sin 2x)}$
$=\frac{2(\cos 2x-\sin 2x)}{(1+\cos 2x)(1+\sin 2x)}$
$=2[\frac{1}{1+\sin 2x}-\frac{1}{1+\cos 2x}]$
$=2[\frac{1}{\alpha }-\frac{1}{\beta }]$