यदि y(alpha)=sqrt{2left(frac{tan alpha+cot al | vedclass

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(A) दिया गया है $y(\alpha)=\sqrt{2\left(\frac{	an \alpha+\cot \alpha}{\sec^2 \alpha}ight)+\csc^2 \alpha}$.चूँकि $1+	an^2 \alpha = \sec^2 \alpha$,ह

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$4$

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(A) दिया गया है $y(\alpha)=\sqrt{2\left(\frac{	an \alpha+\cot \alpha}{\sec^2 \alpha}ight)+\csc^2 \alpha}$.चूँकि $1+	an^2 \alpha = \sec^2 \alpha$,हमारे पास $y(\alpha)=\sqrt{2\left(\frac{\frac{\sin \alpha}{\cos \alpha}+\frac{\cos \alpha}{\sin \alpha}}{\frac{1}{\cos^2 \alpha}}ight)+\csc^2 \alpha}$ है।वर्गमूल के अंदर के व्यंजक को सरल करने पर: $2\left(\frac{\sin^2 \alpha + \cos^2 \alpha}{\sin \alpha \cos \alpha}ight) \cdot \cos^2 \alpha + \csc^2 \alpha = 2\frac{\cos \alpha}{\sin \alpha} + \csc^2 \alpha = 2\cot \alpha + \csc^2 \alpha$.ध्यान दें कि $2\cot \alpha + \csc^2 \alpha = 2\cot \alpha + 1 + \cot^2 \alpha = (1+\cot \alpha)^2$.अतः,$y(\alpha) = \sqrt{(1+\cot \alpha)^2} = |1+\cot \alpha|$.$\alpha \in \left(\frac{3\pi}{4}, \piight)$ के लिए,$\cot \alpha अतः,$y(\alpha) = -(1+\cot \alpha) = -1 - \cot \alpha$.$\alpha$ के सापेक्ष अवकलन करने पर: $\frac{dy}{d\alpha} = -(-\csc^2 \alpha) = \csc^2 \alpha$.$\alpha = \frac{5\pi}{6}$ पर,$\csc \alpha = \csc \frac{5\pi}{6} = 2$.इसलिए,$\frac{dy}{d\alpha} = (2)^2 = 4$.

यदि $y(\alpha)=\sqrt{2\left(\frac{\tan \alpha+\cot \alpha}{1+\tan ^{2} \alpha}\right)+\frac{1}{\sin ^{2} \alpha}}$ जहाँ $\alpha \in\left(\frac{3 \pi}{4}, \pi\right)$ है,तो $\alpha=\frac{5 \pi}{6}$ पर $\frac{d y}{d \alpha}$ ज्ञात कीजिए।

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