tan^2a- tan^2b then prove that sin^2a-sin^2b/(cos^2a)(cos^2b)
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You want to Tan^2a- tan^2b =sin^2a-sin^2b/(cos^2a)(cos^2b)
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Tan^2a- tan^2b =sin^2a-sin^2b/(cos^2a)(cos^2b)
L.H.S = cos^2(a)cos^(b)
=sin^2(a)/cos^2(a) - sin^2(b)/cos^2(b)
= {sin^2(a).cos^2(b) - sin^2(b).cos^2(a)}/cos^2(a)cos^(b)
= {sin^(a).[1 - sin^(b)] - sin^2(b).[1 - sin^2(a)]} /cos^2(a)cos^(b)
={sin^2(a) - sin^(a)sin^(b) - sin^2(b) +sin^2(b)sin^2(a)}/cos^2(a)cos^(b)
=sin^2(a) - sin^2(b) /cos^2(a)cos^(b)//
L.H.S = cos^2(a)cos^(b)
=sin^2(a)/cos^2(a) - sin^2(b)/cos^2(b)
= {sin^2(a).cos^2(b) - sin^2(b).cos^2(a)}/cos^2(a)cos^(b)
= {sin^(a).[1 - sin^(b)] - sin^2(b).[1 - sin^2(a)]} /cos^2(a)cos^(b)
={sin^2(a) - sin^(a)sin^(b) - sin^2(b) +sin^2(b)sin^2(a)}/cos^2(a)cos^(b)
=sin^2(a) - sin^2(b) /cos^2(a)cos^(b)//
very wonderful !!!!!!
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