If y(alpha)=sqrt{2left(frac{tan alpha+cot alp | vedclass

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(A) Given $y(\alpha)=\sqrt{2\left(\frac{	an \alpha+\cot \alpha}{\sec^2 \alpha}ight)+\csc^2 \alpha}$.Since $1+	an^2 \alpha = \sec^2 \alpha$,we have

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$4$

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(A) Given $y(\alpha)=\sqrt{2\left(\frac{	an \alpha+\cot \alpha}{\sec^2 \alpha}ight)+\csc^2 \alpha}$.Since $1+	an^2 \alpha = \sec^2 \alpha$,we have $y(\alpha)=\sqrt{2\left(\frac{\frac{\sin \alpha}{\cos \alpha}+\frac{\cos \alpha}{\sin \alpha}}{\frac{1}{\cos^2 \alpha}}ight)+\csc^2 \alpha}$.Simplifying the expression inside the square root: $2\left(\frac{\sin^2 \alpha + \cos^2 \alpha}{\sin \alpha \cos \alpha}ight) \cdot \cos^2 \alpha + \csc^2 \alpha = 2\frac{\cos \alpha}{\sin \alpha} + \csc^2 \alpha = 2\cot \alpha + \csc^2 \alpha$.Note that $2\cot \alpha + \csc^2 \alpha = 2\cot \alpha + 1 + \cot^2 \alpha = (1+\cot \alpha)^2$.Thus,$y(\alpha) = \sqrt{(1+\cot \alpha)^2} = |1+\cot \alpha|$.For $\alpha \in \left(\frac{3\pi}{4}, \piight)$,$\cot \alpha Therefore,$y(\alpha) = -(1+\cot \alpha) = -1 - \cot \alpha$.Differentiating with respect to $\alpha$: $\frac{dy}{d\alpha} = -(-\csc^2 \alpha) = \csc^2 \alpha$.At $\alpha = \frac{5\pi}{6}$,$\csc \alpha = \csc \frac{5\pi}{6} = 2$.So,$\frac{dy}{d\alpha} = (2)^2 = 4$.

If $y(\alpha)=\sqrt{2\left(\frac{\tan \alpha+\cot \alpha}{1+\tan ^{2} \alpha}\right)+\frac{1}{\sin ^{2} \alpha}}$ for $\alpha \in\left(\frac{3 \pi}{4}, \pi\right)$,then find $\frac{d y}{d \alpha}$ at $\alpha=\frac{5 \pi}{6}$.

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