1. Let $u=\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}$ so that $y=\cot^{-1}(u).$

We need $\dfrac{dy}{dx}$ at $x=\dfrac{5\pi}{6}$.

2. Simplify the expression inside $\cot^{-1}$.

Let $a=\sqrt{1+\sin x},\qquad b=\sqrt{1-\sin x}.$ Then $u=\frac{a+b}{a-b}.$ Multiply numerator and denominator by $a+b$: $u=\frac{(a+b)^2}{a^2-b^2}.$ Now, $a^2-b^2=(1+\sin x)-(1-\sin x)=2\sin x,$ and $(a+b)^2=a^2+b^2+2ab=(1+\sin x)+(1-\sin x)+2\sqrt{(1+\sin x)(1-\sin x)}.$ So, $(a+b)^2=2+2\sqrt{1-\sin^2 x}=2+2|\cos x|.$

Since $x\in\left(\frac{\pi}{2},\pi\right)$, we have $\cos x<0$, hence $|\cos x|=-\cos x.$ Therefore, $(a+b)^2=2-2\cos x=2(1-\cos x).$ Thus, $u=\frac{2(1-\cos x)}{2\sin x}=\frac{1-\cos x}{\sin x}.$ Using the identity $\frac{1-\cos x}{\sin x}=\tan\frac{x}{2},$ we get $u=\tan\frac{x}{2}.$

Hence $y=\cot^{-1}\left(\tan\frac{x}{2}\right).$

3. Use the given interval.

Since $x\in\left(\frac{\pi}{2},\pi\right),$ we have $\frac{x}{2}\in\left(\frac{\pi}{4},\frac{\pi}{2}\right).$ So $\tan\dfrac{x}{2}>0$, and for principal values of $\cot^{-1}$ in $(0,\pi)$, $\cot^{-1}(\tan\theta)=\frac{\pi}{2}-\theta\quad \text{for }\theta\in\left(0,\frac{\pi}{2}\right).$ Taking $\theta=\dfrac{x}{2}$, $y=\frac{\pi}{2}-\frac{x}{2}.$

4. Differentiate: $\frac{dy}{dx}=-\frac{1}{2}.$ Therefore at $x=\frac{5\pi}{6},$ we still have $\left.\frac{dy}{dx}\right|_{x=5\pi/6}=-\frac{1}{2}.$

5. Check options.

• A: $-\dfrac{1}{2}$ ✔
• B: $-1$ ✘
• C: $\dfrac{1}{2}$ ✘
• D: $0$ ✘

So the correct option is A.