Question

If x=secx-cosx, y=sec^nx-cos^nx then [x^2 +4 /y^2+4]× (dy/dx)^2=

Answer 1

Appropriate Question

If

$\rm \: x = sect \: - \: cost$

and

$\rm \: y = {sec}^{n}t - {cos}^{n}t$

$\rm \: \: then \: \: \dfrac{ {x}^{2} + 4}{ {y}^{2} + 4} {\bigg(\dfrac{dy}{dx}\bigg) }^{2} = - - - - - -$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: x = sect \: - \: cost$

So, Consider

$\rm \: {x}^{2} + 4$

$\rm \: = \: {sec}^{2}t + {cos}^{2}t - 2 \times sect \times cost + 4$

$\rm \: = \: {sec}^{2}t + {cos}^{2}t -2 + 4$

$\rm \: = \: {sec}^{2}t + {cos}^{2}t + 2$

$\rm \: = \: {sec}^{2}t + {cos}^{2}t + 2 \times sect \times cost$

$\rm \: = \: {(sect + cost)}^{2}$

$\rm\implies \:\boxed{\tt{ \rm \: {x}^{2} + 4 = \: {(sect + cost)}^{2} }} - - - - (1)$

Also, Given that

$\rm \: y = {sec}^{n}t - {cos}^{n}t$

Now, Consider

$\rm \: {y}^{2} + 4$

$\rm \: = \: {( {sec}^{n}t - {cos}^{n}t)}^{2} + 4$

$\rm \: = \: {sec}^{2n}t + {cos}^{2n}t - 2 \times {sec}^{n}t \times {cos}^{n}t + 4$

$\rm \: = \: {sec}^{2n}t + {cos}^{2n}t - 2 + 4$

$\rm \: = \: {sec}^{2n}t + {cos}^{2n}t + 2$

$\rm \: = \: {( {sec}^{n}t + {cos}^{n}t)}^{2}$

$\rm\implies \:\boxed{\tt{ \rm \: {y}^{2} + 4 = \: {( {sec}^{n}t + {cos}^{n}t)}^{2} \: }} - - - (2)$

Now, Consider

$\rm \: x = sect \: - \: cost$

On differentiating both sides w. r. t. t, we get

$\rm \: \dfrac{dx}{dt} = sect \: tant + sint$

can be further rewritten as

$\rm \: \dfrac{dx}{dt} = sect \: tant + \dfrac{sint}{cost} \times cost$

$\rm \: \dfrac{dx}{dt} = sect \: tant + tant \times cost$

$\rm\implies \:\rm \: \dfrac{dx}{dt} = tant \: (sect + cost) - - - - (3)$

Now, Consider

$\rm \: y = {sec}^{n}t - {cos}^{n}t$

On differentiating both sides w. r. t. t, we get

$\rm \: \dfrac{dy}{dt} = n{sec}^{n - 1}t(sect \: tant) - n{cos}^{n - 1}t( - sint)$

$\rm \: \dfrac{dy}{dt} = n{sec}^{n}t( \: tant) + n{cos}^{n - 1}t(sint)$

can be rewritten as

$\rm \: \dfrac{dy}{dt} = n{sec}^{n}t( \: tant) + n{cos}^{n - 1}t \times \dfrac{sint}{cost} \times cost$

$\rm \: \dfrac{dy}{dt} = n{sec}^{n}t( \: tant) + n{cos}^{n}t \: tant$

$\rm\implies \:\rm \: \dfrac{dy}{dt} = ntant \: ({sec}^{n}t + {cos}^{n}t) - - - (4)$

Now,

$\rm \: \dfrac{dy}{dx} = \dfrac{dy}{dt} \div \dfrac{dx}{dt}$

$\rm \: = \: \dfrac{n \: tant \: ( {sec}^{n}t \: + \: {cos}^{n}t) }{tant(sect \: + \: cost)}$

$\rm \: = \: \dfrac{n\: ( {sec}^{n}t \: + \: {cos}^{n}t) }{sect \: + \: cost}$

So,

$\rm\implies \: {\bigg(\dfrac{dy}{dx}\bigg) }^{2} = \dfrac{ {n}^{2} {( {sec}^{n}t \: + \: {cos}^{n}t) }^{2} }{ {(sect \: + \: cost)}^{2} }$

So, using equation (1) and (2), we get

$\rm \: {\bigg(\dfrac{dy}{dx}\bigg) }^{2} = \dfrac{ {n}^{2} ( {y}^{2} + 4)}{ {x}^{2} + 4}$

$\rm\implies \:\dfrac{ {x}^{2} + 4}{ {y}^{2} + 4} {\bigg(\dfrac{dy}{dx}\bigg) }^{2} = {n}^{2}$

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FORMULA USED

$\boxed{\tt{ \dfrac{d}{dx}secx = secx \: tanx \: }}$

$\boxed{\tt{ \dfrac{d}{dx}cosx \: = \: - \: sinx \: }}$

$\boxed{\tt{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1} \: }}$

$\boxed{\tt{ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \: }}$

$\boxed{\tt{ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \: }}$

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Additional Information

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

Question:-

$\sf If \: x = sec \theta - cos \theta,y = sec {}^{n} \theta - cos {}^{n} \theta, \\ \sf then \bigg( \frac{x {}^{2} + 4 }{y {}^{2} + 4} \bigg) \bigg( \frac{dy}{dx} \bigg) {}^{2} =$

Solution:-

$\sf \large \frac{dy}{dx} = \frac{dy}{ \frac{d \theta}{dx} } = \frac{n(sec {}^{n} \theta \: tan \theta \: + cos {}^{n - 1} \theta \: sin \theta }{sec \theta \: tan \theta \: + sin \theta}$

$\sf \large = \frac{n \: tan \theta(sec {}^{n} \theta + cos {}^{n} \theta)}{tan \theta(sec \theta + cos \theta)}$

$\sf \large( \frac{dy}{dx} ) {}^{2} = \frac{n {}^{2} (sec {}^{n} \theta + cos {}^{n} \theta) {}^{2} }{(sec \theta + cos \theta) {}^{2} }$

$\sf \large \frac{n {}^{2} ((sec {}^{n} \theta - cos {}^{n} \theta) {}^{2} + 4) }{(sec \theta - cos \theta) {}^{2} + 4 } = \frac{n {}^{2}(y {}^{2} + 4) }{(x {}^{2} + 4) }$

Answer:-

${ \boxed{ \sf \large \red{\bigg( \frac{x {}^{2} + 4 }{y {}^{2} + 4} \bigg) \bigg( \frac{dy}{dx} \bigg) {}^{2} = n {}^{2} .}}}$

Hope you have satisfied. ⚘