If tanA=n tanB=msinB then proe that  cos^2A=(m^2-1)/(n^2-1)

SHAIK AASIF AHAMED
9 years ago
Hello student,
I think there is some mistake in your question.Whether it is tanA=msinB or SinA=msinB please make it sure.Please recheck the question and post it correctly so that i can provide you with a meaningful answer.
S. Haris Ahmed Irfan
29 Points
9 years ago
I am sorry

If tanA=n tanB, sinA=msinB then prove that  cos^2A=(m^2-1)/(n^2-1)
SHAIK AASIF AHAMED
9 years ago
Hello student,
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]
Charmish Kaneria
13 Points
5 years ago
sina=msinbor, m=sina/sinb ---------(1) andtana=ntanbor, sina/cosa=n(sinb/cosb)or, n=sinacosb/cosasinbor, n=m (cosb/cosa) -----(2)or, ncosa=mcosbor, n²cos²a=m²cos²bor, n²cos²a=m²(1-sin²b) [∵, sin²a+cos²a=1]or, n²cos²a=m²(1-sin²a/m²) [using (1)]or, n²cos²a=m²{(m²-sin²a)/m²}or, n²cos²a=m²-sin²aor, n²cos²a=m²-(1-cos²a)or, n²cos²a=m²-1+cos²aor, n²cos²a-cos²a=m²-1or, cos²a(n²-1)=m²-1or, cos²a=(m²-1)/(n²-1)
Yash Chourasiya
3 years ago
Dear Student

Please see the solution in the attachment.