If x=sec theta, y=tan theta,then the value of | vedclass

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(C) Given,$x = \sec 	heta$ and $y = 	an 	heta$. First,find the derivatives with respect to $	heta$: $\frac{dx}{d	heta} = \sec 	heta 	an 	heta$

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(C) Given,$x = \sec 	heta$ and $y = 	an 	heta$. First,find the derivatives with respect to $	heta$: $\frac{dx}{d	heta} = \sec 	heta 	an 	heta$ $\frac{dy}{d	heta} = \sec^2 	heta$ Now,find $\frac{dy}{dx}$ using the chain rule: $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{\sec^2 	heta}{\sec 	heta 	an 	heta} = \frac{\sec 	heta}{	an 	heta} = \frac{1/\cos 	heta}{\sin 	heta / \cos 	heta} = \frac{1}{\sin 	heta} = \csc 	heta$. Next,find the second derivative $\frac{d^2y}{dx^2}$: $\frac{d^2y}{dx^2} = \frac{d}{dx}(\csc 	heta) = \frac{d}{d	heta}(\csc 	heta) \cdot \frac{d	heta}{dx}$ $= (-\csc 	heta \cot 	heta) \cdot \frac{1}{\sec 	heta 	an 	heta}$ $= -\frac{1}{\sin 	heta} \cdot \frac{\cos 	heta}{\sin 	heta} \cdot \frac{\cos 	heta}{1} \cdot \frac{\cos 	heta}{\sin 	heta} = -\frac{\cos^3 	heta}{\sin^3 	heta} = -\cot^3 	heta$. At $	heta = \frac{\pi}{4}$: $\frac{d^2y}{dx^2} = -\cot^3(\frac{\pi}{4}) = -(1)^3 = -1$.

If $x=\sec \theta, y=\tan \theta$,then the value of $\frac{d^{2} y}{d x^{2}}$ at $\theta=\frac{\pi}{4}$ is

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