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Prove that \(2cos2A+1\over {2cos2A-1}\) = tan(\(60^{\circ}\) + A)tan(\(60^{\circ}\) – A)

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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 »

Solve : 6 – 10cosx = 3\(sin^2x\)

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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 »

Find general solution of (2sinx – cosx)(1 + cosx) = \(sin^2x\)

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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 »

If \(\lambda x^2 – 10xy + 12y^2 + 5x – 16y – 3\) = 0 represents a pair of straight lines, then \(\lambda\) is equal to

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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 = …

If \(\lambda x^2 – 10xy + 12y^2 + 5x – 16y – 3\) = 0 represents a pair of straight lines, then \(\lambda\) is equal to Read More »

If the straight line 3x + 4y + 5 – k(x + y + 3) = 0 is parallel to y-axis, then the value of k is

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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 …

If the straight line 3x + 4y + 5 – k(x + y + 3) = 0 is parallel to y-axis, then the value of k is Read More »

If x + 4y – 5 = 0 and 4x + ky + 7 = 0 are two perpendicular lines then k is

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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 »

Find the equation of lines which passes through the point (3,4) and the sum of intercepts on the axes is 14.

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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 – …

Find the equation of lines which passes through the point (3,4) and the sum of intercepts on the axes is 14. Read More »

The mean and variance of 5 observations of an experiment are 4 and 5.2 respectively. If from these observations three are 1, 2 and 6, then remaining will be

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Solution : As given \(\bar{x}\) = 4, n = 5 and \({\sigma}^2\) = 5.2. If the remaining observations are \(x_1\), \(x_2\) then \({\sigma}^2\) = \(\sum{(x_i – \bar{x})}^2\over n\) = 5.2 \(\implies\) \({(x_1-4)}^2 + {(x_2-4)}^2 + {(1-4)}^2 + {2-4)}^2 + {(6-4)}^2\over 5\) = 5.2 \(\implies\) \({(x_1-4)}^2 + {(x_2-4)}^2\) = 9  …..(1) Also \(\bar{x}\) = 4 \(\implies\) …

The mean and variance of 5 observations of an experiment are 4 and 5.2 respectively. If from these observations three are 1, 2 and 6, then remaining will be Read More »

A student obtained 75%, 80%, 85% marks in three subjects. If the marks of another subject are added then his average marks can not be less than

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Solution : Total marks obtained from three subjects out of 300 = 75 + 80 + 85 = 240 if the marks of another subject is added then the total marks obtained out of 400 is greater than 240 if marks obtained in fourth subject is 0 then minimum average marks = \(240\over 400\)\(\times\)100 = …

A student obtained 75%, 80%, 85% marks in three subjects. If the marks of another subject are added then his average marks can not be less than Read More »
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