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

We have been given the function

$$y(\alpha)=\sqrt{\,2\left(\dfrac{\tan\alpha+\cot\alpha}{1+\tan^{2}\alpha}\right)+\dfrac{1}{\sin^{2}\alpha}}\,,\qquad\alpha\in\left(\dfrac{3\pi}{4},\pi\right).$$

Our objective is to find $$\dfrac{dy}{d\alpha}$$ at $$\alpha=\dfrac{5\pi}{6}.$$

First we simplify the expression inside the square root. We observe the identity

$$1+\tan^{2}\alpha=\sec^{2}\alpha.$$

Also,

$$\tan\alpha+\cot\alpha =\frac{\sin\alpha}{\cos\alpha}+\frac{\cos\alpha}{\sin\alpha} =\frac{\sin^{2}\alpha+\cos^{2}\alpha}{\sin\alpha\cos\alpha} =\frac{1}{\sin\alpha\cos\alpha}.$$

Substituting these into the fraction we get

$$\dfrac{\tan\alpha+\cot\alpha}{1+\tan^{2}\alpha} =\dfrac{\dfrac{1}{\sin\alpha\cos\alpha}}{\dfrac{1}{\cos^{2}\alpha}} =\dfrac{1}{\sin\alpha\cos\alpha}\cdot\cos^{2}\alpha =\frac{\cos\alpha}{\sin\alpha} =\cot\alpha.$$

So the whole quantity under the square root becomes

$$2\cot\alpha+\frac{1}{\sin^{2}\alpha}.$$

Next we recall the Pythagorean identity

$$\csc^{2}\alpha=1+\cot^{2}\alpha.$$

Using this identity we write

$$2\cot\alpha+\frac{1}{\sin^{2}\alpha} =2\cot\alpha+\csc^{2}\alpha =2\cot\alpha+1+\cot^{2}\alpha =\cot^{2}\alpha+2\cot\alpha+1 =(\cot\alpha+1)^{2}.$$

Therefore

$$y(\alpha)=\sqrt{(\cot\alpha+1)^{2}} =|\cot\alpha+1|.$$

Now we determine the sign of $$\cot\alpha+1$$ in the given interval $$\alpha\in\left(\dfrac{3\pi}{4},\pi\right).$$ In this interval $$\sin\alpha\gt 0$$ and $$\cos\alpha\lt 0,$$ so $$\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}\lt 0.$$ At the left‐end point $$\alpha=\dfrac{3\pi}{4},$$ $$\cot\alpha=-1,$$ making $$\cot\alpha+1=0.$$ As $$\alpha$$ increases beyond $$\dfrac{3\pi}{4},$$ $$\cot\alpha$$ becomes even more negative, hence $$\cot\alpha+1\lt 0$$ throughout the open interval. Thus

$$|\cot\alpha+1|=-(\cot\alpha+1).$$

So we can write, for all $$\alpha\in\left(\dfrac{3\pi}{4},\pi\right),$$

$$y(\alpha)=-(\cot\alpha+1)=-\cot\alpha-1.$$

Now we differentiate. The basic derivative we use is

$$\frac{d}{d\alpha}(\cot\alpha)=-\csc^{2}\alpha.$$

Therefore,

$$\frac{dy}{d\alpha} =-\frac{d}{d\alpha}(\cot\alpha)-\frac{d}{d\alpha}(1) =-(-\csc^{2}\alpha)-0 =\csc^{2}\alpha.$$

We must evaluate this at $$\alpha=\dfrac{5\pi}{6}.$$ We know

$$\sin\left(\dfrac{5\pi}{6}\right)=\frac{1}{2}\quad\Longrightarrow\quad \csc\left(\dfrac{5\pi}{6}\right)=2.$$

Hence

$$\csc^{2}\left(\dfrac{5\pi}{6}\right)=2^{2}=4.$$

Thus,

$$\left.\frac{dy}{d\alpha}\right|_{\alpha=\frac{5\pi}{6}}=4.$$

Hence, the correct answer is Option A.