JEE Main 2020 (Online) 7th January Morning ShiftQuestion 1
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
Differentiation and its Application
Solution
Concept. The expression under the root is $(1+\cot\alpha)^2$; on $\left(\frac{3\pi}{4},\pi\right)$, $1+\cot\alpha<0$. Formulas used. (1) $\dfrac{\tan\alpha+\cot\alpha}{\sec^2\alpha}=\cot\alpha$, $\ \csc^2\alpha=1+\cot^2\alpha$Step 1 - Simplify. By (1), the radicand is $2\cot\alpha+1+\cot^2\alpha=(1+\cot\alpha)^2$. Step 2 - Sign. $\cot\alpha<-1$ on the interval, so $y=-1-\cot\alpha$. Step 3 - Derivative. $y'=\csc^2\alpha$; at $\frac{5\pi}{6}$, $\sin\alpha=\frac12$, so $y'=4$. Answer: (A)