If y(α)=2tan⁡α+cot⁡α1+tan2⁡α+1sin2⁡α,α∈3π4,π, then dydx at α=5π6 is

y(α)=2tan⁡α+cot⁡α1+tan2⁡α+1sin2⁡α=2cos2⁡αsin⁡αcos⁡α+cos⁡αsin⁡α+1sin2⁡α=2cos2⁡αsin2⁡α+cos2⁡αcos⁡αsin⁡α+1sin2⁡α=2sin⁡αcos⁡α+1sin2⁡α=sin⁡α+cos⁡αsin⁡α∴∈3π4,π⇒y(α)=−(sin⁡α+cos⁡α)sin⁡α=−(1+cot⁡α)∴dydα=cosec2⁡α⇒dydxα=5π/6=cosec2⁡5π6=4.

If y(α)=2tan⁡α+cot⁡α1+tan2⁡α+1sin2⁡α,α∈3π4,π, then dydx at α=5π6 is

y(α)=2tan⁡α+cot⁡α1+tan2⁡α+1sin2⁡α=2cos2⁡αsin⁡αcos⁡α+cos⁡αsin⁡α+1sin2⁡α=2cos2⁡αsin2⁡α+cos2⁡αcos⁡αsin⁡α+1sin2⁡α=2sin⁡αcos⁡α+1sin2⁡α=sin⁡α+cos⁡αsin⁡α∴∈3π4,π⇒y(α)=−(sin⁡α+cos⁡α)sin⁡α=−(1+cot⁡α)∴dydα=cosec2⁡α⇒dydxα=5π/6=cosec2⁡5π6=4.

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

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Detailed Solution

$\begin{array}{l} y (α) = \sqrt{2 (\frac{\tan α + \cot α}{1 + \tan^{2} α}) + \frac{1}{\sin^{2} α}} = \sqrt{2 \cos^{2} α (\frac{\sin α}{\cos α} + \frac{\cos α}{\sin α}) + \frac{1}{\sin^{2} α}} \\ = \sqrt{\frac{2 \cos^{2} α (\sin^{2} α + \cos^{2} α)}{\cos αsin α} + \frac{1}{\sin^{2} α}} = \sqrt{\frac{2 \sin αcos α + 1}{\sin^{2} α}} = |\frac{\sin α + \cos α}{\sin α}| \\ ∴\in (\frac{3 π}{4}, π) \Rightarrow y (α) = \frac{- (\sin α + \cos α)}{\sin α} = - (1 + \cot α) \\ ∴ \frac{dy}{dα} = {cosec}^{2} α \Rightarrow {(\frac{dy}{dx})}_{α = 5 π / 6} = {cosec}^{2} (\frac{5 π}{6}) = 4 . \end{array}$