Question

tanA=n tanA
sinA=m sinB
Prove that cos2A=m2-1/n2-1

Answer 1

tan A = n tan B
sin A=m sinB
sin B = sin A / m --------------- (1)
tan A=sinA/cosA
cos A = sin A / tan A = m sin B / n tan B = m cos B / n
cos B = n cos A / m -------------- (2)
squaring and adding (1) and (2)
[sin^2 B + cos^2 B] = sin^2 A / m^2 + n^2 cos^2 A / m^2
1 = [1 - cos^2 A]/m^2 + n^2 cos^2 A / m^2
m^2 = 1 - cos^2 A + n^2 cos^2 A
cos^2 A [n^2 - 1] = [m^2 -1]
cos^2 A = [m^2 -1] / [n^2 - 1]


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Answer 2

sorry but I am able to do this much only

I hope this much help will help you to solve the remaining part