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
Similar Questions
Evaluate sin78 – sin66 – sin42 + sin6.
If A + B + C = \(3\pi\over 2\), then cos2A + cos2B + cos2C is equal to
\(sin5x + sin2x – sinx\over {cos5x + 2cos3x + 2cos^x + cosx}\) is equal to
Find the maximum value of 1 + \(sin({\pi\over 4} + \theta)\) + \(2cos({\pi\over 4} – \theta)\)