The equation of a tangent to the curve y cot | vedclass

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(D) Given the curve $y \cot x = y^3 	an x$. This can be written as $y \cot x - y^3 	an x = 0$,or $y (\cot x - y^2 	an x) = 0$. This implies $y = 0$

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All of the above

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(D) Given the curve $y \cot x = y^3 	an x$. This can be written as $y \cot x - y^3 	an x = 0$,or $y (\cot x - y^2 	an x) = 0$. This implies $y = 0$ or $y^2 = \frac{\cot x}{	an x} = \cot^2 x$. At $x = \frac{\pi}{4}$,$\cot^2(\frac{\pi}{4}) = 1^2 = 1$,so $y^2 = 1$,which gives $y = 1$ or $y = -1$. Case $1$: $y = 1$. Differentiating $y^2 = \cot^2 x$ with respect to $x$,we get $2y \frac{dy}{dx} = 2 \cot x (-\csc^2 x)$. At $x = \frac{\pi}{4}, y = 1$,we have $2(1) \frac{dy}{dx} = 2(1)(-(\sqrt{2})^2) = -4$,so $\frac{dy}{dx} = -2$. The tangent equation is $y - 1 = -2(x - \frac{\pi}{4}) \implies y - 1 = -2x + \frac{\pi}{2} \implies 2x + y = 1 + \frac{\pi}{2} \implies 4x + 2y = 2 + \pi$. Case $2$: $y = -1$. At $x = \frac{\pi}{4}, y = -1$,we have $2(-1) \frac{dy}{dx} = 2(1)(-2) = -4$,so $\frac{dy}{dx} = 2$. The tangent equation is $y + 1 = 2(x - \frac{\pi}{4}) \implies y + 1 = 2x - \frac{\pi}{2} \implies 2x - y = 1 + \frac{\pi}{2} \implies 4x - 2y = 2 + \pi$. Since both $4x + 2y = \pi + 2$ and $4x - 2y = \pi + 2$ are valid tangents,the correct option is $D$.

The equation of a tangent to the curve $y \cot x = y^3 \tan x$ at the point where the abscissa is $\frac{\pi}{4}$ is:

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