# How do you verify sec^2 x - cot^2 ( pi/2-x) =1?

May 5, 2016

see below

#### Explanation:

Use Properties:
$\cos \left(A - B\right) = \cos A \cos B + \sin A \sin B$

$\sin \left(A - B\right) = \sin A \cos B - \cos A \sin B$

Left Side:$= {\sec}^{2} x - {\cot}^{2} \left(\frac{\pi}{2} - x\right)$

$= \frac{1}{\cos} ^ 2 x - {\left[\frac{\cos \left(\left(\frac{\pi}{2}\right) - x\right)}{\sin \left(\left(\frac{\pi}{2}\right) - x\right)}\right]}^{2}$

$= \frac{1}{\cos} ^ 2 x - {\left[\frac{\cos \left(\frac{\pi}{2}\right) \cos x + \sin \left(\frac{\pi}{2}\right) \sin x}{\sin \left(\frac{\pi}{2}\right) \cos x - \cos \left(\frac{\pi}{2}\right) \sin x}\right]}^{2}$

$= \frac{1}{\cos} ^ 2 x - {\left[\frac{0 \cdot \cos x + 1 \cdot \sin x}{1 \cdot \cos x - 0 \cdot \sin x}\right]}^{2}$

$= \frac{1}{\cos} ^ 2 x - {\left[\frac{\sin x}{\cos x}\right]}^{2}$

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

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

$= {\cos}^{2} \frac{x}{\cos} ^ 2 x$

$= 1$

$=$Right Side