Question

if tanA+sinA=m and tanA-sinA=n , then show that m2-n2​

Answer 1

Answer:

If tanA + sinA = m and tanA – sinA = n, show that m2 – n2 = 4√(mn)Solution :- LHS = m2 – n2 = (tanA + sinA)2 – (tanA – sinA)2 = tan2A + sin2A + 2tanAsinA – tan2A – sin2A + 2tanAsinA = 4tanA.sinA RHS = 4√(mn) = 4√(tanA + sinA).(tanA – sinA ) = 4√(tan2A – sin2A = 4√(sin2A/cos2A – sin2A) = 4√{(sin2A – sin2Acos2A)/cos2A} = 4√{sin2A(1 – cos2A)/cos2A} = 4√(sin2A.sin2A/cos2A) = 4(sinA.sinA/cosA) = 4.tanA.sinA = LHS2