How do you prove (cscx + cotx)^2 = (1 + cosx) / (1 - cosx)(cscx+cotx)2=1+cosx1cosx?

2 Answers
Apr 15, 2015

In this way.

The first member is:

(1/sinx+cosx/sinx)^2=(1+cosx)^2/sin^2x=(1+cosx)^2/(1-cos^2x)=(1sinx+cosxsinx)2=(1+cosx)2sin2x=(1+cosx)21cos2x=

(1+cosx)^2/((1+cosx)(1-cosx))=(1+cosx)/(1-cosx)(1+cosx)2(1+cosx)(1cosx)=1+cosx1cosx,

that is the second member.

Apr 15, 2015

(csc x + cotx)^2(cscx+cotx)2

= (1/sin x + cos x/ sin x)^2(1sinx+cosxsinx)2

= (1+cosx)^2 / sin^2 x(1+cosx)2sin2x

= (1+cos x)^2 /(1-cos^2x)(1+cosx)21cos2x

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

= (1+cos x)/(1-cosx)1+cosx1cosx = RHS