Question

TanA=ntanB,sinA=msinB,prove thar cos^2A=m^2-1/n^2-1

Answer 1

here is your answer check it.

Answer 2

Tan A = n Tan B  ---(1)
Sin A = m Sin B  --- (2)
(2) ÷ (1) gives:  CosA = (m/n) CosB    --(3)

From (2)
    Cos² A = 1 - m² Sin²B
                 = 1 - m² [1 - cos²B]
                 = 1 - m² + m² Cos²B
                 = 1 - m² + n² Cos²A      using (3)

=> (n²-1) Cos²A = m² -1 
=>  Cos²A = (m² - 1) / (n² - 1)