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Question
Prove the following identities:
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2A * cos^2B)`
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Solution
L.H.S. = `tan^2A - tan^2B`
= `sin^2A/cos^2A - sin^2B/cos^2B`
= `(sin^2A * cos^2B - sin^2B * cos^2A)/(cos^2A * cos^2B`
= `(sin^2A(1 - sin^2B)-sin^2B(1 - sin^2A))/(cos^2A * cos^2B)`
= `(sin^2A - sin^2A * sin^2B - sin^2B + sin^2A * sin^2B)/(cos^2A * cos^2B`
= `(sin^2A - cancel(sin^2A * sin^2B) - sin^2B + cancel(sin^2A * sin^2B))/(cos^2A * cos^2B`
= `(sin^2A - sin^2B)/(cos^2A * cos^2B)` = R.H.S.
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