(Simple) Trigonometric Identity: $\frac1{\sin x\cos x } = \frac{\sec^2 x}{\tan x}$

Question

Part of a bigger question requires me to prove this trigonometric identity $$\displaystyle \frac{1}{\sin x\cos x } = \frac{\sec^2 x}{\tan x}$$ Can someone shows me how to do it? Thank you

Answer 1

Simplify the right hand side: $$\dfrac{\sec^2(x)}{\tan(x)}=\dfrac{\;\dfrac{1}{\cos^2(x)}\;}{\dfrac{\sin(x)}{\cos(x)}}=\dfrac{\cos(x)}{\sin(x)\cos^2(x)}=\dfrac{1}{\sin(x)\cos(x)}$$