Question

if tan A = m.tan B, prove sin (A-B)/sin(A+B)=m-1/m+1

Answer 1

tanA = m.tanB

tanA/tanB = m

sinA.cosB/cosA.sinB = m

use, componendo or dividendo rule ,

(sinA.cosB + cosA.sinB)/(sinA.cosB-cosA.sinB) = (m+1)/(m-1)

[ use, sin(A - B) = sinA.cosB-cisA.sinB
sin(A + B) = sinA.cosB+ cosA.sinB ]

sin(A +B)/sin(A - B) = (m+1)/(m-1)

hence,
sin(A -B)/sin(A + B) = (m -1)/( m +1)

Answer 2

tanA/tanB = m

=>

sinA.cosB/cosA.sinB = m

use, componendo or dividendo rule ,

(sinA.cosB + cosA.sinB)/(sinA.cosB-cosA.sinB) = (m+1)/(m-1)

[ use, sin(A - B) = sinA.cosB-cisA.sinB

sin(A + B) = sinA.cosB+ cosA.sinB ]

=>

sin(A +B)/sin(A - B) = (m+1)/(m-1)

hence,

sin(A -B)/sin(A + B) = (m -1)/( m +1)