Cos^2y-sin^2x/sin^2y-cos^2x=-1 , can you prove it?

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Answer 1 · 2018-03-24T11:28:36

#(Cos^2y-sin^2x)/(sin^2y-cos^2x)=-1 #

Mr. Noah G. suggested two option for your question.The proof of third option is given below.

#(Cos^2y-sin^2x)/(sin^2y-cos^2x)=-1 #

Here,

#LHS=(Cos^2y-sin^2x)/(sin^2y-cos^2x)#

#=((color(red)(1-sin^2y))-(color(red)(1-cos^2x)))/(sin^2y-cos^2x)#...#to(cos^2theta+sin^2theta=1)#

#=(cancel1-sin^2y-cancel1+cos^2x)/(sin^2y-cos^2x)#

#=(-sin^2y+cos^2x)/(sin^2y-cos^2x)#

#=-[(sin^2y-cos^2x)/(sin^2y-cos^2x)]#

#=-1#

#=RHS#