Question #58780

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Answer 1 · 2017-11-01T00:48:07

#RHS=(sin^2x)/(2+sin2xcscx)#

#=(2sin(x/2)cos(x/2))^2/(2+2sinxcosx xx1/sinx)#

#=(4sin^2(x/2)cos^2(x/2))/(2(1+cosx ))#

#=(4sin^2(x/2)cos^2(x/2))/(2*2cos^2(x /2))#

#=sin^2(x/2)=LHS#

Answer 2 · 2017-11-01T17:18:01

Formulas to be used in the proof.

Right Hand Side:

#sin^2x/(2+sin2xcscx)=sin^2x/(2+2sinx cosx cscx)#

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

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

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

#=((1-cosx)cancel(1+cosx))/(2cancel(1+cosx ))#

#=(1-cosx)/2#

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

#:.=#Left Hand Side