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