# Question #b2d25

Mar 28, 2017

$\frac{1 - {\cot}^{2} x}{\cot x - {\sec}^{2} x} = \frac{2 \cos 2 x}{2 - \sin 2 x}$

#### Explanation:

$\frac{1 - {\cot}^{2} x}{\cot x - {\sec}^{2} x} = \frac{1 - {\left(\cos \frac{x}{\sin} x\right)}^{2}}{\left(\cos \frac{x}{\sin} x\right) - {\left(\frac{1}{\sin} x\right)}^{2}}$

$\text{ } = \frac{1 - {\cos}^{2} \frac{x}{\sin} ^ 2 x}{\cos \frac{x}{\sin} x - \frac{1}{\sin} ^ 2 x}$

$\text{ } = \frac{\frac{{\sin}^{2} x - {\cos}^{2} x}{\sin} ^ 2 x}{\frac{\cos x \sin x - 1}{\sin} ^ 2 x}$

$\text{ } = \frac{{\sin}^{2} x - {\cos}^{2} x}{\cos x \sin x - 1}$

$\text{ } = \frac{2}{2} \cdot \frac{{\cos}^{2} x - {\sin}^{2} x}{1 - \cos x \sin x}$

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

$\text{ } = 2 \cdot \frac{\cos 2 x}{2 - \sin 2 x}$

That is about as simple as I can get the expression