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

##### 1 Answer
Nov 21, 2015

We will be using the following properties:

1. $\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$ (definition of secant)
2. $\cot \left(x\right) = \cos \frac{x}{\sin} \left(x\right)$ (definition of cotangent)
3. $\cos \left(- x\right) = \cos \left(x\right)$ (cosine is even)
4. $\cos \left(x - \frac{\pi}{2}\right) = \sin \left(x\right)$
5. ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$

By 1, we have

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

By 3 and 4,

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

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

By 5,

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

Thus, by 2,

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

Substituting back then gives us

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

$\therefore {\sec}^{2} \left(\frac{\pi}{2} - x\right) - 1 = {\cot}^{2} \left(x\right) \text{ Q.E.D.}$