if tanA +sinA= m and tanA-sinA=n, then prove that (m²-n²)=4 whole under root mn
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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
i hope this will help u
and if u have any problem with this .... then u can ask .....
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
i hope this will help u
and if u have any problem with this .... then u can ask .....
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