Solution : Since PS is the median, so S is the mid point of triangle PQR. So, Coordinates of S = (\({7+6\over 2}, {3 – 1\over 2}\)) = (\(13\over 2\), 1) Slope of line PS = (1 – 2)/(13/2 – 2) = \(-2\over 9\) Required equation passes through (1, -1) is y + 1 = […]
Solution : We have [\(\vec{a}\) + \(\vec{b}\) \(\vec{b}\) + \(\vec{c}\) \(\vec{c}\) + \(\vec{a}\)] = {(\(\vec{a}\) + \(\vec{b}\))\(\times\)(\(\vec{b}\) + \(\vec{c}\))}.(\(\vec{c}\) + \(\vec{a}\)) = {\(\vec{a}\)\(\times\)\(\vec{b}\) + \(\vec{a}\)\(\times\)\(\vec{c}\) + \(\vec{b}\)\(\times\)\(\vec{b}\) + \(\vec{b}\)\(\times\)\(\vec{c}\)}.(\(\vec{c}\) + \(\vec{a}\)) {\(\vec{b}\)\(\times\)\(\vec{b}\) = 0} = {\(\vec{a}\)\(\times\)\(\vec{b}\) + \(\vec{a}\)\(\times\)\(\vec{c}\) + \(\vec{b}\)\(\times\)\(\vec{c}\)}.(\(\vec{c}\) + \(\vec{a}\)) = (\(\vec{a}\times\vec{b}\)).\(\vec{c}\) + (\(\vec{a}\times\vec{c}\)).\(\vec{c}\) + (\(\vec{b}\times\vec{c}\)).\(\vec{c}\) + (\(\vec{a}\times\vec{b}\)).\(\vec{a}\) + (\(\vec{a}\times\vec{c}\)).\(\vec{a}\) + (\(\vec{b}\times\vec{c}\)).\(\vec{a}\) =
Solution : \(\vec{a}\times\vec{b}\) = \(\vec{c}\) and \(\vec{b}\times\vec{c}\) = \(\vec{a}\) \(\implies\) \(\vec{c}\perp\vec{a}\) , \(\vec{c}\perp\vec{b}\) and \(\vec{a}\perp\vec{b}\), \(\vec{a}\perp\vec{c}\) \(\implies\) \(\vec{a}\perp\vec{b}\), \(\vec{b}\perp\vec{c}\) and \(\vec{c}\perp\vec{a}\) \(\implies\) \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are mutually perpendicular vectors. Again, \(\vec{a}\times\vec{b}\) = \(\vec{c}\) and \(\vec{b}\times\vec{c}\) = \(\vec{a}\) \(\implies\) |\(\vec{a}\times\vec{b}\)| = |\(\vec{c}\)| and |\(\vec{b}\times\vec{c}\)| = |\(\vec{a}\)| \(\implies\) \(|\vec{a}||\vec{b}|sin{\pi\over 2}\) = |\(\vec{c}\)| and \(|\vec{b}||\vec{c}|sin{\pi\over 2}\) = |\(\vec{a}\)|
Solution : Unit vectors perpendicular to \(\vec{a}\) & \(\vec{b}\) = \(\pm\)\(\vec{a}\times\vec{b}\over |\vec{a}\times\vec{b}|\) \(\therefore\) \(\vec{a}\times\vec{b}\) = \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 1 & 2 & -2 \\ \end{vmatrix}\) = \(-5\hat{i} – 5\hat{j} – 5\hat{k}\) \(\therefore\) Unit Vectors = \(\pm\) \(-5\hat{i} – 5\hat{j} – 5\hat{k}\over 5\sqrt{3}\) Hence the required
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
Solve : cos3x + sin2x – sin4x = 0 Read More »
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
Solve : 6 – 10cosx = 3\(sin^2x\) Read More »
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 »
