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Question
Prove the following identity:
tan2A − sin2A = tan2A · sin2A
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Solution
tan2A − sin2A = tan2A · sin2A
LHS:
tan2A − sin2A
We know that:
`tan A = (sin A)/(cos A) => tan^2A = (sin^2 A)/(cos^2 A)`
`tan^2 A - sin^2A = (sin^2 A)/(cos^2 A) - sin^2 A`
`= (sin^2 A - sin^2 A cos^2 A)/(cos^2 A)`
Factor out sin2A:
= `(sin^2 A(1 - cos^2 A))/(cos^2 A)`
1 − cos2A = sin2A
= `(sin^2 A * sin^2 A)/(cos^2 A) = (sin^4 A)/(cos^2 A)`
RHS:
tan2A − sin2A = tan2A · sin2A
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