Proving This Trigonometric Identity

[Ace.]

1. Prove:$$\frac{cscx +cotx}{cscx-cotx} = \frac{1+2cosx+cos^2x}{sin^2x}$$

Homework Equations

tanx = $\frac{sinx}{cosx}$
cotx = $\frac{cosx}{sinx}$
cscx
secx
cotx
$sin^2x + cos^2x = 1$

The Attempt at a Solution

Left side:
=$\frac{cscx +cotx}{cscx-cotx}$
=$\frac{1/sinx + cosx / sinx}{1/sinx - cosx/sinx}$
=$\frac{1+cosx}{1-cosx}$

 

[Infrared]

You are almost there! What could you multiply the numerator and denominator of your fraction to get the RHS?

 

[Saitama]

You are almost there. What is sin^2x in terms of cos^2x?

EDIT: Just a minute late. -.-

 

[Ace.]

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

thanks guys