Question

if tan A =n tanB
sin A =m sin B
prove Cos²A =m²-1/n²-1​

Answer 1

Answer:

given

tanA = ntanB and sinA = msinB

squaring...

sin^2A = m^2 sin^2B

sin^2B = sin^2A/m^2

cos^2B = 1 - sin^2A/m^2

now

tan^2A = n^2 tan^2B

tan^2B = tan^2A/n^2

sin^2B/cos^2B = sin^2A/n^2cos^2A

putting the values of sin^2B and cos^2B, we have

(sin^2A/m^2)/(1-sin^2A/m^2) = sin^2A/n^2cos^2A

1/(m^2 - sin^2A) = 1/n^2cos^2A

n^2cos^2A = m^2 - sin^2A

n^2cos^2A = m^2 - 1 + cos^2A

n^2cos^2A - cos^2A = m^2 - 1

cos^2A (n^2 - 1) = m^2 - 1

cos^2A = (m^2 - 1)/(n^2 - 1)

proved