Maths Questions - Page 50 of 52

Prove that : \(cos^{-1}{12\over 13}\) + \(sin^{-1}{3\over 5}\) = \(sin^{-1}{56\over 65}\)

Solution : We have, L.H.S. = \(cos^{-1}{12\over 13}\) + \(sin^{-1}{3\over 5}\) = \(tan^{-1}{5\over 12}\) + \(tan^{-1}{3\over 4}\) \(\because\) [ \(cos^{-1}{12\over 13}\) = \(tan^{-1}{5\over 12}\) & \(sin^{-1}{3\over 5}\) = \(tan^{-1}{3\over 4}\) ] L.H.S. = \(tan^{-1}({{{5\over 12} + {3\over 4}}\over {1 – {5\over 12}.{3\over 4}}})\) = \(tan^{-1}{56\over 33}\) R.H.S. = \(sin^{-1}{56\over 65}\) = \(tan^{-1}{56\over 33}\) L.H.S = …

Prove that : \(cos^{-1}{12\over 13}\) + \(sin^{-1}{3\over 5}\) = \(sin^{-1}{56\over 65}\) Read More »

Find the value of \(sin^{-1}({-\sqrt{3}\over 2})\) + \(cos^{-1}(cos({7\pi\over 6}))\).

Solution : \(sin^{-1}({-\sqrt{3}\over 2})\) = – \(sin^{-1}({\sqrt{3}\over 2})\) = \(-\pi\over 3\) \(cos^{-1}(cos({7\pi\over 6}))\) = \(cos^{-1}(cos({2\pi – {5\pi\over 6}}))\) = \(cos^{-1}(cos({5\pi\over 6}))\) = \(5\pi\over 6\) Hence \(sin^{-1}({-\sqrt{3}\over 2})\) + \(cos^{-1}(cos({7\pi\over 6}))\) = \(-\pi\over 3\) + \(5\pi\over 6\) = \(\pi\over 2\) Similar Questions Solve the equation : 2\(tan^{-1}({2x+1})\) = \(cos^{-1}x\) Prove that : \(sin^{-1}{12\over 13}\) + \(cot^{-1}{4\over …

Find the value of \(sin^{-1}({-\sqrt{3}\over 2})\) + \(cos^{-1}(cos({7\pi\over 6}))\). Read More »

Prove that \(\int_{0}^{\pi/2}\) log(sinx)dx = \(\int_{0}^{\pi/2}\) log(cosx)dx = -\(\pi\over 2\)log 2.

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Solution : Let I = \(\int_{0}^{\pi/2}\) log(sinx)dx …….(i) then I = \(\int_{0}^{\pi/2}\) \(log(sin({\pi\over 2}-x))\)dx = \(\int_{0}^{\pi/2}\) log(cosx)dx …….(ii) Adding (i) and (ii), we get 2I = \(\int_{0}^{\pi/2}\) log(sinx)dx + \(\int_{0}^{\pi/2}\) log(cosx)dx = \(\int_{0}^{\pi/2}\) (log(sinx)dx + log(cosx)) \(\implies\) \(\int_{0}^{\pi/2}\) log(sinxcosx)dx = \(\int_{0}^{\pi/2}\) \(log({2sinxcosx\over 2})\)dx = \(\int_{0}^{\pi/2}\) \(log({sin2x\over 2})\)dx = \(\int_{0}^{\pi/2}\) log(sin2x)dx – \(\int_{0}^{\pi/2}\) log(2)dx …

Prove that \(\int_{0}^{\pi/2}\) log(sinx)dx = \(\int_{0}^{\pi/2}\) log(cosx)dx = -\(\pi\over 2\)log 2. Read More »

Evaluate : \(\int\) \(cos^4xdx\over {sin^3x{(sin^5x + cos^5x)^{3\over 5}}}\)

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Solution : I = \(\int\) \(cos^4xdx\over {sin^3x{(sin^5x + cos^5x)^{3\over 5}}}\) = \(\int\) \(cos^4xdx\over {sin^6x{(1 + cot^5x)^{3\over 5}}}\) = \(\int\) \(cot^4xcosec^2xdx\over {{(1 + cot^5x)^{3\over 5}}}\) Put \(1+cot^5x\) = t \(5cot^4xcosec^2x\)dx = -dt = -\(1\over 5\) \(\int\) \(dt\over {t^{3/5}}\) = -\(1\over 2\) \(t^{2/5}\) + C = -\(1\over 2\) \({(1+cot^5x)}^{2/5}\) + C Similar Questions What is the integration …

Evaluate : \(\int\) \(cos^4xdx\over {sin^3x{(sin^5x + cos^5x)^{3\over 5}}}\) Read More »

Evaluate : \(\int\) \(dx\over {3sinx + 4cosx}\)

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Solution : I = \(\int\) \(dx\over {3sinx + 4cosx}\) = \(\int\) \(dx\over {3[{2tan{x\over 2}\over {1+tan^2{x\over 2}}}] + 4[{1-tan^2{x\over 2}\over {1+tan^2{x\over 2}}}]}\) = \(\int\) \(sec^2{x\over 2}dx\over {4+6tan{x\over 2}-4tan^2{x\over 2}}\) let \(tan{x\over 2}\) = t, \(\therefore\) \({1\over 2}sec^2{x\over 2}\)dx = dt so I = \(\int\) \(2dt\over {4+6t-4t^2}\) = \(1\over 2\) \(\int\) \(dt\over {1-(t^2-{3\over 2}t})\) = \(1\over 2\) …

Evaluate : \(\int\) \(dx\over {3sinx + 4cosx}\) Read More »

Find the asymptotes of the hyperbola \(2x^2 + 5xy + 2y^2 + 4x + 5y\) = 0. Find also the general equation of all the hyperbolas having the same set of asymptotes.

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Solution : Let \(2x^2 + 5xy + 2y^2 + 4x + 5y + k\) = 0 be asymptotes. This will represent two straight line so \(abc + 2fgh – af^2 – bg^2 – ch^2\) = 0 \(\implies\) 4k + 25 – \(25\over 2\) – 8 – \(25\over 4\)k = 0 \(\implies\) k = 2 \(\implies\) …

Find the asymptotes of the hyperbola \(2x^2 + 5xy + 2y^2 + 4x + 5y\) = 0. Find also the general equation of all the hyperbolas having the same set of asymptotes. Read More »

Find the equation of the tangent to the hyperbola \(x^2 – 4y^2\) = 36 which is perpendicular to the line x – y + 4 = 0

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Solution : Let m be the slope of the tangent, since the tangent is perpendicular to the line x – y = 0 \(\therefore\) m\(\times\)1 = -1 \(\implies\) m = -1 Since \(x^2-4y^2\) = 36 or \(x^2\over 36\) – \(y^2\over 9\) = 1 Comparing this with \(x^2\over a^2\) – \(y^2\over b^2\) = 1 \(\therefore\); \(a^2\) …

Find the equation of the tangent to the hyperbola \(x^2 – 4y^2\) = 36 which is perpendicular to the line x – y + 4 = 0 Read More »

The eccentricity of the conjugate hyperbola to the hyperbola \(x^2-3y^2\) = 1 is

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Solution : Equation of the conjugate hyperbola to the hyperbola \(x^2-3y^2\) = 1 is \(-x^2-3y^2\) = 1 \(\implies\) \(-x^2\over 1\) + \(y^2\over {1/3}\) = 1 Here \(a^2\) = 1, \(b^2\) = \(1\over 3\) \(\therefore\) eccentricity e = \(\sqrt{1 + a^2/b^2}\) = \(\sqrt{1+3}\) = 2 Similar Questions Angle between asymptotes of hyperbola xy=8 is Find the …

The eccentricity of the conjugate hyperbola to the hyperbola \(x^2-3y^2\) = 1 is Read More »

If the foci of a hyperbola are foci of the ellipse \(x^2\over 25\) + \(y^2\over 9\) = 1. If the eccentricity of the hyperbola be 2, then its equation is :

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Solution : For ellipse e = \(4\over 5\), so foci = (\(\pm\)4, 0) for hyperbola e = 2, so a = \(ae\over e\) = \(4\over 2\) = 2, b = \(2\sqrt{4-1}\) = 2\(\sqrt{3}\) Hence the equation of the hyperbola is \(x^2\over 4\) – \(y^2\over 12\) = 1 Similar Questions Find the equation of the ellipse …

If the foci of a hyperbola are foci of the ellipse \(x^2\over 25\) + \(y^2\over 9\) = 1. If the eccentricity of the hyperbola be 2, then its equation is : Read More »

Find the period of the function f(x) = \(e^{x-[x]+|cos\pi x|+|cos2\pi x|+ ….. + |cosn\pi x|}\)

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Solution : f(x) = \(e^{x-[x]+|cos\pi x|+|cos2\pi x|+ ….. + |cosn\pi x|}\) Period of x – [x] = 1 Period of \(|cos\pi x|\) = 1 Period of \(|cos2\pi x|\) = \(1\over 2\) ………………………………. Period of \(|cosn\pi x|\) = \(1\over n\) So period of f(x) will be L.C.M of all period = 1. Similar Questions If y …

Find the period of the function f(x) = \(e^{x-[x]+|cos\pi x|+|cos2\pi x|+ ….. + |cosn\pi x|}\) Read More »