How do you prove #(cscx + cotx)^2 = (1 + cosx) / (1 - cosx)#?

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Answer 1 · 2015-04-15T15:28:32

In this way.

The first member is:

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

#(1+cosx)^2/((1+cosx)(1-cosx))=(1+cosx)/(1-cosx)#,

that is the second member.

Answer 2 · 2015-04-15T15:37:22

#(csc x + cotx)^2#

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

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

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

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

= # (1+cos x)/(1-cosx)# = RHS