Solution : The value of Cot 0 degrees is equal to \(\infty\) or not defined. Proof : \(\angle\) A is made smaller and smaller in the right \(\Delta\) ABC till it becomes zero. As \(\angle\) A gets smaller and smaller the length of the side BC decreases. The point C gets closer to the point B …
Solution : The value of Tan 0 degrees is equal to 0. Proof : \(\angle\) A is made smaller and smaller in the right \(\Delta\) ABC till it becomes zero. As \(\angle\) A gets smaller and smaller the length of the side BC decreases. The point C gets closer to the point B and finally when …
Solution : The value of Cos 0 degrees is equal to 1. Proof : \(\angle\) A is made smaller and smaller in the right \(\Delta\) ABC till it becomes zero. As \(\angle\) A gets smaller and smaller the length of the side BC decreases. The point C gets closer to the point B and finally when …
Solution : The value of sin 0 degrees is equal to 0. Proof : \(\angle\) A is made smaller and smaller in the right \(\Delta\) ABC till it becomes zero. As \(\angle\) A gets smaller and smaller the length of the side BC decreases. The point C gets closer to the point B and finally …
Solution : We have 1 + \(sin({\pi\over 4} + \theta)\) + \(2cos({\pi\over 4} – \theta)\) = 1 + \(1\over sqrt{2}\)\((cos\theta + cos\theta)\) + \(\sqrt{2}\)\((cos\theta + cos\theta)\) = 1 + \(({1\over \sqrt{2}} + \sqrt{2})\) + \((cos\theta + cos\theta)\) = 1 + \(({1\over \sqrt{2}} + \sqrt{2})\).\(\sqrt{2}cos(\theta – {pi\over 4})\) \(\therefore\) Maximum Value = 1 + \(({1\over \sqrt{2}} …
Solution : The expression = (sin78 – sin42) – (sin66 – sin6) = 2cos(60)sin(18) – 2cos36.sin30 = sin18 – cos36 = \(({\sqrt{5} – 1\over 4})\) – \(({\sqrt{5} + 1\over 4})\) = -\(1\over 2\) Similar Questions Find the maximum value of 1 + \(sin({\pi\over 4} + \theta)\) + \(2cos({\pi\over 4} – \theta)\) If A + B …
Solution : cos2A + cos2B + cos2C = 2cos(A+B)cos(A-B) + cos2C = 2cos(\(3\pi\over 2\) – C)cos(A-B) + cos2C \(\because\) A + B + C = \(3\pi\over 2\) = -2sinC cos(A-B) + 1 – 2\(sin^2C\) = 1 – 2sinC[cos(A-B)+sinC] = 1 – 2sinC[cos(A-B) + sin(\(3\pi\over 2\)-(A+B))] = 1 – 2sinC[cos(A-B)-cos(A+B)] = 1 – 4sinA sinB sinC …
If A + B + C = \(3\pi\over 2\), then cos2A + cos2B + cos2C is equal to Read More »
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 …
\(sin5x + sin2x – sinx\over {cos5x + 2cos3x + 2cos^x + cosx}\) is equal to Read More »
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 + …
Prove that \(2cos2A+1\over {2cos2A-1}\) = tan(\(60^{\circ}\) + A)tan(\(60^{\circ}\) – A) Read More »
