Prove that #csc4A+csc8A=cot2A-cot8A#?

1 Answer
P dilip_k
Oct 27, 2017

#RHS=cot2A-cot8A#

#=(cos2A)/(sin2A)-(cos8A)/(sin8A)#

#=(cos2Asin8A-cos8Asin2A)/(sin2Asin8A)#

#=sin(8A-2A)/(sin2Asin8A)#

#=(2cos2Asin6A)/(2cos2Asin2Asin8A)#

#=(sin8A+sin4A)/(sin4Asin8A)#

#=(sin8A)/(sin4Asin8A)+(sin4A)/(sin4Asin8A)#

#=1/(sin4A)+1/(sin8A)#

#=csc4A+csc8A=LHS#

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