Question

if x = sin^3t / √cost , y = cos^3t / √cost
then prove that dy/dx = - cot 3t ​

Answer 1

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$\bf \: if \tt \: {x \: = \frac{ {sin}^{3} t}{ \sqrt{cos \: 2t} } \: \: \: or \: \: \: y = \: \frac{ {cos}^{3}t }{ \sqrt{cos \: 2t} } } \\ \\ \bf \: then \: prove \: that \tt \: \frac{dy}{dx} = - cot \: 3t$

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$\bold{\huge{\underline{\underline{Used\:Formulas}}}}$

$\boxed{1 . \sf \frac{d}{dx}( \frac{p}{q} ) = \frac{q \times \frac{dp}{dx}- p \times \frac{dq}{dx}}{ {q}^{2} } } \\ \boxed{2 . \sf \: 2sin \: a \times \: cosb = sin(a + b) + sin(a - b)} \\ \boxed{3 . \sf \: sin3a = 3sina + {4sin}^{3} a} \\ \boxed{4 . \sf \frac{cos \: \theta}{sin \: \theta} = cot \theta } \\ \boxed{5 . \sf \frac{d}{dx}sinx = cosx } \\ \boxed{ 6 .\sf \: \frac{d}{dx}cosx = - sinx }$

$\bold{\huge{\underline{Solution-}}}$

we need to find dy/dx = -cot3t

$\tt \: \frac{dy}{dx} = \frac{ (\frac{dy}{dt} )}{ (\frac{dx}{dt}) }$

$\implies \tt \: \frac{ \frac{d}{dt} \frac{ {cos}^{3}t }{ \sqrt{cos2 \: t} } }{ \frac{d}{dt} \frac{ {sin}^{3}t }{ \sqrt{cos2 \: t} } }$

$\implies \large \tt \: \: \frac{ \{ \: \frac{ \sqrt{cos2 \: t} + \frac{d}{dt}(cos \: t) {}^{3} - {cos}^{3} t \times \frac{d}{dt}(cos \: 2t) {}^{ \frac{1}{2} } }{( \sqrt{cos \: 2t}) {}^{2} } \}}{ \{ \: \frac{ \sqrt{cos \: 2t} \times \frac{d}{dt} (sin \: t) {}^{3} - {sin}^{3} t \times \frac{d}{dt}(cos \: 2t) {}^{ \frac{1}{2} } }{( \sqrt{cos2t}) {}^{2} } \}}$

$\implies \large \tt \: \{ \: \frac{ \sqrt{cos \: 2t} \times 3(cos \: t) {}^{3 - 1} \times ( - sin \: t) - ( {cos}^{3}t) \times \frac{1}{2}(cos \: 2t) {}^{ \frac{1}{2} - 1 } \times \frac{d}{dt}cos \: 2t }{ \sqrt{cos \: 2t} \times 3(sin \: t) {}^{3 - 1} \times cos \: t - {sin}^{3}t \: \times \frac{1}{2} ( \: cos2 \: t) {}^{ \frac{1}{2} - 1} \times \frac{d}{dt}cos2 \: t } \}$

$\implies \large \tt \: \frac{ {- 3 \sqrt{cos \: 2t} \times {cos}^{2}t \times sint \: - \frac{ {cos}^{3} t}{2 \sqrt{cos \: 2t} } \times ( - sin \: 2t) \times 2 \times 1 }{}}{ {3 \sqrt{cos \: 2t} \times sin {}^{2} t \times cost - \frac{ {sin}^{3}t }{2 \sqrt{cos \: 2t} } \times ( - sin2t) \times 2 \times 1 }{} }$

$\implies \large \tt \: \frac{cos {}^{2}t \{ \: \frac{ - 3cos \: 2t \times sint + cos \: t \: \times \: sin \: 2t}{ \sqrt{cos \: 2t} } \} }{ {sin}^{2}t \{ \: \frac{3 \: cos \: 2t \times cost + sin \: t \times sin2t}{ \sqrt{cos 2t\: } } \} }$

$\implies \large \tt \: \frac{ {cos}^{2}t }{sin {}^{2}t } \{ \: \frac{ - 3 \times 2sint \times cos2t + 2sin \: 2t \times cost}{3 \times 2 \: cos2t \: cost \: + 2 \: sin \: t \: sin2t} \}$

$\implies \large \tt \: \frac{ {cos}^{2}t }{ {sin}^{2}t } \: \frac{[ - 3 \{ \: sin(t + 2t) + sin(t - 2t) \} + \{sin(2t + t) +sin (2t - t) \}]}{ \: [3 \times \{ \: cos(2t + t) + cos(2t - t) \} + \{ \: cos(t - 2t) - cos(t + 2t) \}]}$

$\implies \large \tt \: \frac{ {cos}^{2} t}{ {sin}^{2}t } \frac{ \:[ - 3sin \: 3t \: - 3sin( - t) + sin \: 3t + sint] }{ \:[3cos \: 3t \: + 3 \: cost + cos( - t) - cos \: 3t]}$

$\implies \large \tt \: \frac{ {cos}^{2} t}{ {sin}^{2} t} [\frac{ - 2 \: sin3t \: + 4sin \: t}{2cos \: 3t + 4 \: cos \: t} ]$

$\implies \large \tt \: \frac{ {cos}^{2}t }{ {sin}^{2} t} [ \frac{ - 2 \times \{ \: 3sin \: t - 4sin {}^{3}t \} + 4 \: sint }{2 \times \{ \: 4cos {}^{3}t \: - 3cos \: t \} \: + 4 \: cost } ]$

$\implies \large \tt \: \frac{ {cos}^{2}t }{ {sin}^{2}t } [ \frac{ - 2 \: sint + 8 {sin}^{3}t }{8 {cos}^{3}t - 2cost } ]$

$\implies \large \tt \: \frac{2 {cos}^{2}t \times sin \: t \{ - 1 + 4 {sin}^{2}t \} }{2 {sin}^{2}t \: cost \{4 {cos}^{2}t - 1 \} }$

$\implies \large \tt \: \frac{cos \: t}{sin \: t} [ \frac{ - 1 + 4(1 - {cos}^{2}t) }{4(1 - {sin}^{2}t) - 1 } ]$

$\implies \large \tt \: \frac{cos \: t}{sin \: t} \frac{ [3 - 4 {cos}^{2}t ] }{[3 - 4 {sin}^{2}t ]}$

$\implies \large \tt \: \frac{ - [4 {cos}^{3}t - 3cost ]}{[3sin \: t - 4 {sin}^{3}t ]}$

$\implies \large \tt \: \frac{ - cos3t}{sin3t} \\ \\ \implies \large \tt \: \frac{dy}{dt} = - cot3t$