Question

pls solve this question correctly and give me step by step explanation​

Answer 1

Step-by-step explanation:

We have,

$x = t + \cot(t)$ and $y=\sin(t)$

Now,

$\frac{dx}{dt} = 1 - \cosec^{2} (t)$

And,

$\frac{dy}{dt} = \cos(t)$

Now,

$\frac{dy}{dx} = \frac{ \dfrac{dy}{dt} }{ \dfrac{dx}{dt} } = \frac{ \cos(t) }{1 - \cosec^{2} (t) }$

$\implies\frac{dy}{dx} = \frac{ \cos(t) \sin ^{2} (t) }{ \sin^{2} (t) - 1 }$

$\implies\frac{dy}{dx} = \frac{ \cos(t) \sin ^{2} (t) }{ - \cos^{2} (t) }$

$\implies\frac{dy}{dx} = \frac{ \sin ^{2} (t) }{ - \cos (t) }$

$\implies\frac{dy}{dx} = \frac{ \sin (t) }{ - \cos (t) } . \sin(t)$

$\implies\frac{dy}{dx} = - \tan(t) . \sin(t)$

$\implies\frac{d^{2} y}{dx^{2} } = - \frac{d}{dx} \{\tan(t) . \sin(t) \}$

$\implies\frac{d^{2} y}{dx^{2} } = - \{\sec^{2} (t) . \sin(t) + \tan(t) \cos(t) \}. \frac{dt}{dx}$

$\implies\frac{d^{2} y}{dx^{2} } = - \{\sec^{2} (t) . \sin(t) + \sin(t) \}. \frac{1}{1 - \cosec ^{2} (t) }$

$\implies\frac{d^{2} y}{dx^{2} } = - \frac{\sec^{2} (t) . \sin(t) + \sin(t) }{1 - \cosec ^{2} (t) }$

$\implies\frac{d^{2} y}{dx^{2} } = - \frac{ \sin(t) \{\sec^{2} (t) + 1 \} }{1 - \cosec ^{2} (t) }$

$\implies\frac{d^{2} y}{dx^{2} } = - \frac{ \sin(t). \sin^{2} (t) \{\sec^{2} (t) + 1 \} }{ \sin ^{2} (t) - 1}$

$\implies\frac{d^{2} y}{dx^{2} } = \frac{ \sin(t). \sin^{2} (t) \{\sec^{2} (t) + 1 \} }{1 - \sin ^{2} (t) }$

$\implies\frac{d^{2} y}{dx^{2} } = \frac{ \sin(t). \sin^{2} (t) \{\sec^{2} (t) + 1 \} }{ \cos^{2} (t) }$

$\implies\frac{d^{2} y}{dx^{2} } = \frac{ \sin(t). \sin^{2} (t) \{1 + \cos^{2} (t) \} }{ \cos^{4} (t) }$

$\implies\frac{d^{2} y}{dx^{2} } = \frac{\sin^{3} (t) \{1 + \cos^{2} (t) \} }{ \cos^{4} (t) }$

$\implies\frac{d^{2} y}{dx^{2} } = \tan^{3} (t) \sec(t) \{1 + \cos^{2} (t) \}$

$\implies\frac{d^{2} y}{dx^{2} } = \tan^{3} (t) \sec(t) + \tan ^{3} (t) \cos(t) \}$