Question

if tanA +sinA= m and tanA-sinA=n, then prove that (m²-n²)=4 whole under root mn

Answer 1

L.H.S =
tan^2A + sin^2A + 2tanAsinA - tan^2A - sin^2A + 2tanAsinA = 4tanAsinA

R.H.S =
4√mn = 4√(tanA + sinA) ( tanA - sinA)
4√tan^2A - sin^2A
4√sin^2A/cos^2A - sin^2A
4√(sin^2A- sin^2Acos^2A)/ cos^2A
4√sin^2A(1-cos^2A)/cos^2A
4√tan^2A. sin^2A
4tanAsinA

L.H.S =R.H.S
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