How do you verify #(1+cotalpha)^2-2cotalpha=1/((1-cosalpha)(1+cosalpha))#?

2 Answers
Noah G
Dec 6, 2016

Start by transforming all the terms into sine and cosine using the identity #color(red)(cot theta = 1/tantheta = 1/(sintheta/costheta) = costheta/sintheta#

#(1 + cosalpha/sinalpha)^2 - (2cosalpha)/sinalpha = 1/((1 - cosalpha)(1 + cosalpha))#

#1 + (2cosalpha)/sinalpha + cos^2alpha/sin^2alpha - (2cosalpha)/sinalpha = 1/(1- cos^2alpha)#

The #(2cosalpha)/sinalpha#'s cancel each other out

#1 + cos^2alpha/sin^2alpha = 1/(1 - cos^2alpha)#

We use the identity #color(red)(sin^2beta + cos^2beta =1-> sin^2beta = 1- cos^2beta# at this point in the process.

#(sin^2alpha + cos^2alpha)/sin^2alpha = 1/sin^2alpha#

#1/sin^2alpha = 1/sin^2alpha#

#LHS = RHS#

Identity Proved!

Hopefully this helps!

P dilip_k
Dec 6, 2016

#LHS=(1+cotalpha)^2-2cotalpha#

#=1^2+cot^2alpha+2cotalpha-2cotalpha#

#=csc^2alpha#

#=1/sin^2alpha#

#=1/(1-cos^2alpha)#

#=1/((1-cosalpha)(1+cosalpha))=RHS#

Proved

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