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Question
Prove the following:
`(2cos2"A" + 1)/(2cos2"A" - 1)` = tan(60° + A) tan(60° − A)
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Solution
R.H.S. = tan(60° + A) tan(60° − A)
= `(sin(60^circ + "A")sin(60^circ - "A"))/(cos(60^circ + "A")cos(60^circ - "A")`
= `(2sin(60^circ + "A")sin(60^circ - "A"))/(2cos(60^circ + "A")cos(60^circ - "A")`
= `(cos[60^circ + "A" - (60^circ - "A")] - cos(60^circ + "A" + 60^circ - "A"))/(cos(60^circ + "A" + 60^circ - "A") + cos[60^circ + "A" - (60^circ - "A")]`
= `(cos2"A" - cos120^circ)/(cos120^circ - cos2"A")`
= `(cos2"A" - cos(180^circ - 60^circ))/(cos(180^circ - 60^circ) + cos2"A")`
= `(cos2"A" - (- cos 60^circ))/(- cos60^circ + cos2"A")`
= `(cos2"A" + 1/2)/(-1/2 + cos2"A")`
= `(2cos2"A" + 1)/(2cos2"A" - 1)`
= L.H.S.
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