# How to prove that cos^2x-sin^2x/sinxcosx=2-sec^2x/tanx?

Mar 5, 2018

See explanation

#### Explanation:

We want to verify the identity

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

We will use the pythagorean trigonometric identity

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

$R H S = \frac{2 - {\sec}^{2} \left(x\right)}{\tan} \left(x\right)$

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

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

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

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

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