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

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(A) આપેલ છે $y(\alpha)=\sqrt{2\left(\frac{	an \alpha+\cot \alpha}{\sec^2 \alpha}ight)+\csc^2 \alpha}$.$1+	an^2 \alpha = \sec^2 \alpha$ હોવાથી,$y(\

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$4$

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(A) આપેલ છે $y(\alpha)=\sqrt{2\left(\frac{	an \alpha+\cot \alpha}{\sec^2 \alpha}ight)+\csc^2 \alpha}$.$1+	an^2 \alpha = \sec^2 \alpha$ હોવાથી,$y(\alpha)=\sqrt{2\left(\frac{\frac{\sin \alpha}{\cos \alpha}+\frac{\cos \alpha}{\sin \alpha}}{\frac{1}{\cos^2 \alpha}}ight)+\csc^2 \alpha}$.વર્ગમૂળની અંદરના પદનું સાદું રૂપ આપતા: $2\left(\frac{\sin^2 \alpha + \cos^2 \alpha}{\sin \alpha \cos \alpha}ight) \cdot \cos^2 \alpha + \csc^2 \alpha = 2\frac{\cos \alpha}{\sin \alpha} + \csc^2 \alpha = 2\cot \alpha + \csc^2 \alpha$.નોંધો કે $2\cot \alpha + \csc^2 \alpha = 2\cot \alpha + 1 + \cot^2 \alpha = (1+\cot \alpha)^2$.તેથી,$y(\alpha) = \sqrt{(1+\cot \alpha)^2} = |1+\cot \alpha|$.$\alpha \in \left(\frac{3\pi}{4}, \piight)$ માટે,$\cot \alpha તેથી,$y(\alpha) = -(1+\cot \alpha) = -1 - \cot \alpha$.$\alpha$ ની સાપેક્ષ વિકલન કરતા: $\frac{dy}{d\alpha} = -(-\csc^2 \alpha) = \csc^2 \alpha$.$\alpha = \frac{5\pi}{6}$ આગળ,$\csc \alpha = \csc \frac{5\pi}{6} = 2$.તેથી,$\frac{dy}{d\alpha} = (2)^2 = 4$.

જો $y(\alpha)=\sqrt{2\left(\frac{\tan \alpha+\cot \alpha}{1+\tan ^{2} \alpha}\right)+\frac{1}{\sin ^{2} \alpha}}$ જ્યાં $\alpha \in\left(\frac{3 \pi}{4}, \pi\right)$ હોય,તો $\alpha=\frac{5 \pi}{6}$ આગળ $\frac{d y}{d \alpha}$ શોધો.

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