# If m = tanA + sinA  n = tanA – sinAProve thatm2-n2 = 4(mn)^1/2

Jitender Singh IIT Delhi
10 years ago













Saloni sharma
14 Points
6 years ago
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 = LHS
Saloni sharma
14 Points
6 years ago
. If tanA + sinA = m and tanA – sinA = n, show that m2 – n2 = 4√(mn)Sol - 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 = LHS
Krish Gupta
Hello student, please find answer to your question.The science lies in basic algebra andd then playing with equations.