Question

If tana=n tanb and sina=m sinb ,prove that cos 2 a=m 2 -1/n n -1

Answer 1

sina=msinb
or, m=sina/sinb ---------(1)  and
tana=ntanb
or, sina/cosa=n(sinb/cosb)
or, n=sinacosb/cosasinb
or, n=m (cosb/cosa) -----(2)
or, ncosa=mcosb
or, n²cos²a=m²cos²b
or, n²cos²a=m²(1-sin²b) [∵, sin²a+cos²a=1]
or, n²cos²a=m²(1-sin²a/m²) [using (1)]
or, n²cos²a=m²{(m²-sin²a)/m²}
or, n²cos²a=m²-sin²a
or, n²cos²a=m²-(1-cos²a)
or, n²cos²a=m²-1+cos²a
or, n²cos²a-cos²a=m²-1
or, cos²a(n²-1)=m²-1
or, cos²a=(m²-1)/(n²-1)

Answer 2

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]