Prove (csc x + cot x)/(csc x - cot x) = (1 + 2cos x + cos²x)/sin²x

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1. Prove:[tex]\frac{cscx +cotx}{cscx-cotx} = \frac{1+2cosx+cos^2x}{sin^2x}[/tex]

Homework Equations


tanx = [itex]\frac{sinx}{cosx}[/itex]
cotx = [itex]\frac{cosx}{sinx}[/itex]
cscx
secx
cotx
[itex]sin^2x + cos^2x = 1[/itex]

The Attempt at a Solution



Left side:
=[itex]\frac{cscx +cotx}{cscx-cotx}[/itex]
=[itex]\frac{1/sinx + cosx / sinx}{1/sinx - cosx/sinx}[/itex]
=[itex]\frac{1+cosx}{1-cosx}[/itex]
 
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Discussion

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You are almost there! What could you multiply the numerator and denominator of your fraction to get the RHS?
 
You are almost there. What is sin^2x in terms of cos^2x?

EDIT: Just a minute late. -.-
 
=[itex]\frac{1+cosx}{1-cosx}[/itex]
=[itex]\frac{1+cosx}{1-cosx}[/itex] × [itex]\frac{1+cosx}{1+cosx}[/itex]
= [itex]\frac{1+2cosx + cos^2x}{1-cos^2x}[/itex]
=[itex]\frac{1+2cosx+cos^2x}{sin^2x}[/itex]

thanks guys