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

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