Let 0

Question

(B) The given equation is $(1+\sin x)y^3 - (\cos x)y^2 + 2(1+\sin x)y - 2\cos x = 0$.Rearranging the terms,we get $(1+\sin x)y(y^2+2) - \cos x(y^2+2)

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$-\frac{1}{2}$

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(B) The given equation is $(1+\sin x)y^3 - (\cos x)y^2 + 2(1+\sin x)y - 2\cos x = 0$.Rearranging the terms,we get $(1+\sin x)y(y^2+2) - \cos x(y^2+2) = 0$.This simplifies to $(y^2+2)((1+\sin x)y - \cos x) = 0$.Since $y^2+2 = 0$ has no real solutions for $y$,we must have $(1+\sin x)y - \cos x = 0$,which implies $y = \frac{\cos x}{1+\sin x}$.Using the half-angle formulas $\cos x = \frac{1-	an^2(x/2)}{1+	an^2(x/2)}$ and $\sin x = \frac{2	an(x/2)}{1+	an^2(x/2)}$,let $t = 	an(x/2)$.Then $y = \frac{1-t^2}{1+t^2} / (1 + \frac{2t}{1+t^2}) = \frac{1-t^2}{1+t^2+2t} = \frac{(1-t)(1+t)}{(1+t)^2} = \frac{1-t}{1+t}$.We want to find $\frac{dy}{dt}$ at $x = \frac{\pi}{2}$,which means $t = 	an(\frac{\pi}{4}) = 1$.$y = \frac{1-t}{1+t} \Rightarrow \frac{dy}{dt} = \frac{-(1+t) - (1-t)}{(1+t)^2} = \frac{-2}{(1+t)^2}$.At $t = 1$,$\frac{dy}{dt} = \frac{-2}{(1+1)^2} = \frac{-2}{4} = -\frac{1}{2}$.

Let $0 < x < \pi$ and $y(x)$ be given by $(1+\sin x)y^3 - (\cos x)y^2 + 2(1+\sin x)y - 2\cos x = 0$. The derivative of $y$ with respect to $\tan \frac{x}{2}$ at $x = \frac{\pi}{2}$ is:

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