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If m = tanA + sinA n = tanA – sinA Prove that m 2 -n 2 = 4(mn)^1/2

If m = tanA + sinA
  n = tanA – sinA
Prove that
m2-n2 = 4(mn)^1/2  

Grade:11

4 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
8 years ago
Ans:Hello student, please find answer to your question
m =tanA + sinA
n =tanA - sinA
m^{2}-n^{2} =(tanA + sinA)^{2} - (tanA-sinA)^{2}
m^{2}-n^{2} =4tanA.sinA
mn = (tanA+sinA)(tanA-sinA)
mn = tan^{2}A-sin^{2}A
mn = sin^{2}A(\frac{1}{cos^{2}A}-1)
mn = sin^{2}A(\frac{1-cos^{2}A}{cos^{2}A})
mn = sin^{2}A(\frac{sin^{2}A}{cos^{2}A})
mn = sin^{2}A.tan^{2}A
\sqrt{mn} = sinA.tanA
m^{2}-n^{2} =4tanA.sinA
m^{2}-n^{2} =4\sqrt{mn}
Saloni sharma
14 Points
5 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
5 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
askIITians Faculty 82 Points
2 years ago
m^{2}-n^{2} =4\sqrt{mn}m^{2}-n^{2} =4tanA.sinA\sqrt{mn} = sinA.tanAmn = sin^{2}A.tan^{2}Amn = sin^{2}A(\frac{sin^{2}A}{cos^{2}A})mn = sin^{2}A(\frac{1-cos^{2}A}{cos^{2}A})mn = sin^{2}A(\frac{1}{cos^{2}A}-1)mn = tan^{2}A-sin^{2}Amn = (tanA+sinA)(tanA-sinA)m^{2}-n^{2} =4tanA.sinAm^{2}-n^{2} =(tanA + sinA)^{2} - (tanA-sinA)^{2}n =tanA - sinAm =tanA + sinAHello student, please find answer to your question.The science lies in basic algebra andd then playing with equations.

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