How do you verify the identity #(csctheta-cottheta)^2=(1-costheta)/(1+costheta)#?

Question

5074 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-21T08:03:24

the identity is verified

Since #csctheta=1/sintheta and cottheta=costheta/(sintheta)#,

you can substitute and have:

#(csctheta-cottheta)^2=(1/sintheta-costheta/(sintheta))^2#

#=((1-costheta)/(sintheta))^2=(1-costheta)^2/(sin^2theta)#

Since #sin^2theta=1-cos^2theta#, the expression becomes:

#=(1-costheta)^2/(1-cos^2theta)=(1-costheta)^cancel2/(cancel((1-costheta))(1+costheta))=(1-costheta)/(1+costheta)#

and the identity is verified