Question

if tana =ntanb and sina=msinb , prove that cos^2a =m^2-1/n^2-1​

Answer 1

Answer:

m^2-1/n^2-1

=[(sin^2 a/sin^2 b)-1 ]/[(tan^2 a/ tan^2 b)-1]

=[(sin^2 a -sin^2 b)/sin^2 b]/[(tan^2 a-tan^2 b)/tan^2 b]

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