Solution :
L.H.S. = \(2sin2xcos3x + sin2x\over{2cos3x.cos2x + 2cos3x + 2cos^2x}\)
= \(sin2x[2cos3x+1]\over {2[cos3x(cos2x+1)+(cos^2x)]}\)
= \(sin2x[2cos3x+1]\over {2[cos3x(2cos^2x)+(cos^2x)]}\)
= \(sin2x[2cos3x+1]\over {2cos^2x(2cos3x+1)}\) = tanx
Similar Questions
Evaluate sin78 – sin66 – sin42 + sin6.
If A + B + C = \(3\pi\over 2\), then cos2A + cos2B + cos2C is equal to
Find the maximum value of 1 + \(sin({\pi\over 4} + \theta)\) + \(2cos({\pi\over 4} – \theta)\)
Prove that \(2cos2A+1\over {2cos2A-1}\) = tan(\(60^{\circ}\) + A)tan(\(60^{\circ}\) – A)
