If yα=2tanα+cotα1+tan2α+1sin2α , α∈3π4,π, then dydαat α=5π6is

y=2cos2αsinαcosα+1sin2α=2cotα+cosec2α=1+cotα=−1−cotα⇒dydα=cosec2α ⇒dydαα=5π6=cosec2αα=5π6=4

If yα=2tanα+cotα1+tan2α+1sin2α , α∈3π4,π, then dydαat α=5π6is

y=2cos2αsinαcosα+1sin2α=2cotα+cosec2α=1+cotα=−1−cotα⇒dydα=cosec2α ⇒dydαα=5π6=cosec2αα=5π6=4

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If $y (α) = \sqrt{2 [\frac{\tan α + \cot α}{1 + \tan^{2} α}] + \frac{1}{\sin^{2} α}}, α \in [\frac{3 π}{4}, π]$ , then $\frac{d y}{d α}$ at $α = \frac{5 π}{6}$ is

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$- \frac{1}{4}$

$\frac{4}{3}$

$- 4$

answer is A.

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$y = \sqrt{\frac{2 \cos^{2} α}{\sin α \cos α} + \frac{1}{\sin^{2} α}} = \sqrt{2 \cot α + \cos e c^{2} α} = |1 + \cot α| = - 1 - \cot α$

$\Rightarrow \frac{d y}{d α} = \cos e c^{2} α \Rightarrow {(\frac{d y}{d α})}_{α = \frac{5 π}{6}} = {(\cos e c^{2} α)}_{α = \frac{5 π}{6}} = 4$

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If yα=2tanα+cotα1+tan2α+1sin2α , α∈3π4,π, then dydαat α=5π6is

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If $y (α) = \sqrt{2 [\frac{\tan α + \cot α}{1 + \tan^{2} α}] + \frac{1}{\sin^{2} α}}, α \in [\frac{3 π}{4}, π]$ , then $\frac{d y}{d α}$ at $α = \frac{5 π}{6}$ is