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Question
Prove that `(sin "A" - 2sin^3 "A")/(2cos^3 "A" - cos "A") = tan "A"`
Solution
Given
`(sin "A" - 2sin^2 "A")/(2cos^2 "A" - cos "A") = tan "A"`
L.H.S = `(sin "A" - 2 sin^3 "A")/(2 cos^3 "A" - cos "A")`
`= (sin "A" (1 - 2sin^2 "A"))/(cos "A" (2cos^2 "A" - 1))``
`= (sin"A"(sin^2 "A" + cos^2 "A" - 2sin^2 "A"))/(cos"A"(2cos^2"A" - sin^2 "A" - cos^2 "A"))`
`=(sin"A"(cos^2 "A" - sin^2 "A"))/(cos"A"(cos^2 "A" - sin^2 "A"))`
= `sin"A"/cos"A"`
= tan A
= R.H.S
Hence proved.
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