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

Feb 5, 2016

Using the following:

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

We have

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

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

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

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

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

$= {\cos}^{2} \frac{x}{\cos} ^ 2 \left(x\right)$

$= 1$