Prove that \(2cos2A+1\over {2cos2A-1}\) = tan(\(60^{\circ}\) + A)tan(\(60^{\circ}\) – A)

Solution :

R.H.S. = tan(\(60^{\circ}\) + A)tan(\(60^{\circ}\) – A)

= (\(tan60^{\circ}+tanA\over {1-tan60^{\circ}tanA}\))(\(tan60^{\circ}-tanA\over {1+tan60^{\circ}tanA}\))

= (\(\sqrt{3}+tanA\over {1-\sqrt{3}tanA}\))(\(\sqrt{3}-tanA\over {1+\sqrt{3}tanA}\))

= \(3-tan^2A\over{1-3tan^2A}\) = \(3cos^2A-sin^2A\over {cos^2A-3sin^2A}\)

= \(2cos^2A+cos^2A-2sin^2A+sin^2A\over {2cos^2A-2sin^2A-sin^2A-cos^2A}\)

= \(2(cos^2A-sin^2A)+cos^2A+sin^2A\over {2(cos^2A-sin^2A)-(sin^2A+cos^2A)}\)

= \(2cos2A+1\over {2cos2A-1}\) = L.H.S


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