Question

prove that m^2 - n^2 = 4√mn

Answer 1

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} 4 years ago

Approved If tanA + sinA = m and tanA – sinA = n, show that m2 – n2 = 4√(mn)