How do you verify the identity #(csc x - cot x)^2 = (1 - cos x)/(1+cosx)#?

1 Answer
Hriman
Mar 6, 2018

See below

Explanation:

#(csc-cotx)^2=(1-cosx)/(1+cosx)#
#(csc^2x-2cotxcscx+cot^2x)=(1-cosx)/(1+cosx)#
#(1/sin^2x-(2cosx)/sin^2x+cos^2x/sin^2x)=(1-cosx)/(1+cosx)#

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

#(((1-cosx)(1-cosx))/(1-cos^2x))=(1-cosx)/(1+cosx)#

#((cancel(1-cosx)(1-cosx))/(cancel(1-cosx)(1+cosx)))=(1-cosx)/(1+cosx)#

#(1-cosx)/(1+cosx)=(1-cosx)/(1+cosx)#

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