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 »
Solution : we have, cos3x + (sin2x – sin4x) = 0 = cos3x – 2sinx.cos3x = 0 \(\implies\) (cos3x)(1 – 2sinx) = 0 \(\implies\) cos3x = 0 or sinx = \(1\over 2\) \(\implies\) cos3x = 0 = cos\(\pi\over 2\) or sinx = \(1\over 2\) = sin\(\pi\over 6\) \(\implies\) 3x = 2n\(\pi\) \(\pm\) \(\pi\over 2\) or …
Solution : we have, 6 – 10cosx = 3\(sin^2x\) \(\therefore\) 6 – 10cosx = 3 – 3\(cos^2x\) \(\implies\) 3\(cos^2x\) – 10cosx + 3 = 0 \(\implies\) (3cosx-1)(cosx-3) = 0 \(\implies\) cosx = \(1\over 3\) or cosx = 3 Since cosx = 3 is not possible as -1 \(\le\) cosx \(\le\) 1 \(\therefore\) cosx = \(1\over …
Solution : (2sinx – cosx)(1 + cosx) – (1 – \(cos^2x\)) = 0 \(\therefore\) (1 + cosx)(2sinx – cosx – 1 + cosx) = 0 \(\therefore\) (1 + cosx)(2sinx – 1) = 0 \(\implies\) cosx = -1 or sinx = \(1\over 2\) \(\implies\) cosx = -1 = cos\(\pi\) \(\implies\) x = 2n\(\pi\) + \(\pi\) = …
Find general solution of (2sinx – cosx)(1 + cosx) = \(sin^2x\) Read More »
Solution : Comparing with \(ax^2+2hxy+by^2+2gx+2fy+c\) = 0 Here a = \(\lambda\), b = 12, c = -3, f = -8, g = 5/2, h = -5 Using condition \(abc+2fgh-af^2-bg^2-ch^2\) = 0, we have \(\lambda\)(12)(-3) + 2(-8)(5/2)(-5) – \(\lambda\)(64) – 12(25/4) + 3(25) = 0 \(\implies\) -36\(\lambda\) + 200 – 64\(\lambda\) – 75 + 75 = …
Solution : A straight line is parallel to y-axis, if its y-coefficient is zero i.e. 4 – k = 0 i.e. k = 4 Similar Questions The slope of tangent parallel to the chord joining the points (2, -3) and (3, 4) is If the line 2x + y = k passes through the point …
Solution : \(m_1\) = -\(1\over 4\) \(m_2\) = -\(4\over k\) Two lines are perpendicular if \(m_1 m_2\) = -1 \(\implies\) (-\(1\over 4\))\(\times\)(-\(4\over k\)) = -1 \(\implies\) k = -1 Similar Questions If the straight line 3x + 4y + 5 – k(x + y + 3) = 0 is parallel to y-axis, then the value …
If x + 4y – 5 = 0 and 4x + ky + 7 = 0 are two perpendicular lines then k is Read More »
Solution : Let the equation of line be \(x\over a\) + \(y\over b\) = 1 …..(i) This line passes through (3,4), therefore \(3\over a\) + \(4\over b\) = 1 …….(ii) It is given that a + b = 14 \(\implies\) b = 14 – a in (ii), we get \(3\over a\) + \(4\over 14 – …
