If $$y(x) = \cot^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)$$, $$x \in \left(\frac{\pi}{2}, \pi\right)$$, then $$\frac{dy}{dx}$$ at $$x = \frac{5\pi}{6}$$ is:

We have been given

$$y(x)=\cot^{-1}\!\left(\dfrac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right),\qquad x\in\left(\dfrac{\pi}{2},\pi\right).$$

Put

$$A=\sqrt{1+\sin x}$$, $$\qquad B=\sqrt{1-\sin x}$$

so that

$$y=\cot^{-1}\!\left(\dfrac{A+B}{A-B}\right).$$

Let

$$R(x)=\dfrac{A+B}{A-B}.$$ Then $$y=\cot^{-1}(R).$$

To find $$\dfrac{dy}{dx}$$ we first differentiate $$R(x).$$

Since $$A=(1+\sin x)^{1/2},$$ by the chain rule

$$\dfrac{dA}{dx}=\dfrac{1}{2}(1+\sin x)^{-1/2}\cdot\cos x=\dfrac{\cos x}{2\sqrt{1+\sin x}}=\dfrac{\cos x}{2A}.$$

Similarly, $$B=(1-\sin x)^{1/2},$$ so

$$\dfrac{dB}{dx}=\dfrac{1}{2}(1-\sin x)^{-1/2}\cdot(-\cos x)=\dfrac{-\cos x}{2\sqrt{1-\sin x}}=\dfrac{-\cos x}{2B}.$$

Write the numerator and denominator of $$R$$ as

$$N=A+B,\qquad D=A-B,$$ so $$R=\dfrac{N}{D}.$$

Using the quotient rule, $$\dfrac{dR}{dx}=\dfrac{D\dfrac{dN}{dx}-N\dfrac{dD}{dx}}{D^{2}},$$ where $$\dfrac{dN}{dx}= \dfrac{dA}{dx}+\dfrac{dB}{dx},\qquad \dfrac{dD}{dx}= \dfrac{dA}{dx}-\dfrac{dB}{dx}.$$

Now we evaluate every quantity at $$x=\dfrac{5\pi}{6}.$$ At this point $$\sin\!\left(\dfrac{5\pi}{6}\right)=\dfrac12,\qquad \cos\!\left(\dfrac{5\pi}{6}\right)=-\dfrac{\sqrt3}{2}.$$

Therefore

$$A_{0}=A\Big|_{x=5\pi/6}=\sqrt{1+\dfrac12}=\sqrt{\dfrac32} =\dfrac{\sqrt3}{\sqrt2},$$ $$B_{0}=B\Big|_{x=5\pi/6}=\sqrt{1-\dfrac12}=\sqrt{\dfrac12} =\dfrac1{\sqrt2}.$$

The derivatives become

$$\dfrac{dA}{dx}\Bigg|_{x=5\pi/6}= \dfrac{\cos x}{2A}\Bigg|_{x=5\pi/6}= \dfrac{-\dfrac{\sqrt3}{2}}{2\dfrac{\sqrt3}{\sqrt2}} =-\dfrac{\sqrt2}{4},$$

$$\dfrac{dB}{dx}\Bigg|_{x=5\pi/6}= \dfrac{-\cos x}{2B}\Bigg|_{x=5\pi/6}= \dfrac{+\dfrac{\sqrt3}{2}}{2\dfrac1{\sqrt2}} =\dfrac{\sqrt3}{2\sqrt2}=\dfrac{\sqrt6}{4}.$$

Hence

$$N_{0}=A_{0}+B_{0}=\dfrac{\sqrt3+1}{\sqrt2},\qquad D_{0}=A_{0}-B_{0}=\dfrac{\sqrt3-1}{\sqrt2},$$

$$\dfrac{dN}{dx}\Bigg|_{0}= -\dfrac{\sqrt2}{4}+\dfrac{\sqrt6}{4} =\dfrac{-\sqrt2+\sqrt6}{4},$$

$$\dfrac{dD}{dx}\Bigg|_{0}= -\dfrac{\sqrt2}{4}-\dfrac{\sqrt6}{4} =\dfrac{-\sqrt2-\sqrt6}{4}.$$

The derivative of $$R$$ at the point is

$$\begin{aligned} \dfrac{dR}{dx}\Bigg|_{0} &=\dfrac{D_{0}\big(\dfrac{dN}{dx}\big)_{0}-N_{0}\big(\dfrac{dD}{dx}\big)_{0}} {D_{0}^{2}}\\[4pt] &=\dfrac{\dfrac{\sqrt3-1}{\sqrt2}\cdot\dfrac{-\sqrt2+\sqrt6}{4} -\dfrac{\sqrt3+1}{\sqrt2}\cdot\dfrac{-\sqrt2-\sqrt6}{4}} {\left(\dfrac{\sqrt3-1}{\sqrt2}\right)^{2}}\\[6pt] &=\dfrac{2}{\;2-\sqrt3\;}. \end{aligned}$$

Next we need $$\dfrac{dy}{dx}.$$ The standard derivative formula for the inverse cotangent is

$$\dfrac{d}{dx}\bigl[\cot^{-1}u\bigr]=-\dfrac{u'}{1+u^{2}}.$$

Here $$u=R(x),$$ so

$$\dfrac{dy}{dx}=-\dfrac{\dfrac{dR}{dx}}{1+R^{2}}.$$ At $$x=\dfrac{5\pi}{6}$$ we already have $$\dfrac{dR}{dx}=\dfrac{2}{2-\sqrt3}.$$

We still need $$R_{0}=R\bigl(\tfrac{5\pi}{6}\bigr).$$ Using $$R_{0}=\dfrac{A_{0}+B_{0}}{A_{0}-B_{0}} =\dfrac{\sqrt3+1}{\sqrt3-1} =2+\sqrt3,$$ so

$$1+R_{0}^{2}=1+(2+\sqrt3)^{2} =1+\bigl(4+4\sqrt3+3\bigr) =8+4\sqrt3 =4(2+\sqrt3).$$

Consequently

$$\dfrac{dy}{dx}\Bigg|_{x=5\pi/6} =-\dfrac{\dfrac{2}{2-\sqrt3}}{4(2+\sqrt3)} =-\dfrac{2}{4(2-\sqrt3)(2+\sqrt3)}.$$

Because $$(2-\sqrt3)(2+\sqrt3)=4-3=1,$$ the expression simplifies to

$$\dfrac{dy}{dx}\Bigg|_{x=5\pi/6} =-\dfrac{2}{4} =-\dfrac12.$$

Hence, the correct answer is Option C.