Tan(x/2)cos^2(x)-tan(x/2) = sin(x)/sec(x)-sin(x) Verify?

Question

Work one side only to verify

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Answer 1 · 2017-11-18T06:13:52

For proof, please refer to the explanation below.

#RHS=sinx/secx-sinx#

#=sinx(1/secx-1)#

#=sinx(cosx-1)#

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

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

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

#=tan(x/2)*[(1+cosx)(cosx-1)]#

#=tan(x/2)*(cos^2x-1)#

#=tan(x/2)*cos^2x-tan(x/2)#

#=LHS#

Answer 2 · 2017-11-18T09:26:08

#LHS=tan(x/2)cos^2(x)-tan(x/2) #

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

#=-tan(x/2)(sin^2(x)) #

#=-tan(x/2)(2sin(x/2)cos(x/2))^2 #

#=-sin(x/2)/cos(x/2)(2cos^2(x/2))*2sin^2(x/2) #

#=-2sin(x/2)cos(x/2)(1-cos(x)) #

#=-sin(x)(1-1/sec(x)) #

#= sin(x)/sec(x)-sin(x) =RHS#