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

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Differentiation

y (α) = 2 (\frac{t a n α + c o t α}{1 + t a n ^{2} α}) + \frac{1}{s i n ^{2} α}, α \in (\frac{3 π}{4}, π)

then

\frac{d y}{d α}

α = \frac{5 π}{6}

is:

Answer: (A)

Simplify the radicand: with

1 + tan^{2} α = sec^{2} α

and

tan α + cot α = \frac{1}{s i n α c o s α}

, the first term becomes

2 cot α

, so the expression under the root is

2 cot α + csc^{2} α = 1 + 2 cot α + cot^{2} α = (1 + cot α)^{2}

. Hence

y = ∣1 + cot α ∣

, and on

(\frac{3 π}{4}, π)

where

cot α < - 1

this is

y = - (1 + cot α)

, so

\frac{d y}{d α} = csc^{2} α

. At

α = \frac{5 π}{6}

sin α = \frac{1}{2}

, giving

\frac{d y}{d α} = 4

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Single correct option