Consider $$f(x) = \tan^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)$$, $$x \in \left(0, \frac{\pi}{2}\right)$$. A normal to $$y = f(x)$$ at $$x = \frac{\pi}{6}$$ also passes through the point

We have the function

$$y=f(x)=\tan^{-1}\!\left(\sqrt{\dfrac{1+\sin x}{1-\sin x}}\right),\qquad x\in\left(0,\dfrac{\pi}{2}\right).$$

For differentiation it is convenient to remove the square root first. Let us set

$$u(x)=\sqrt{\dfrac{1+\sin x}{1-\sin x}}.$$

Inside the square root we see the ratio $$\dfrac{1+\sin x}{1-\sin x}.$$ Multiplying the numerator and the denominator by $$1+\sin x$$ (so that the denominator becomes $$1-\sin^{2}x$$) gives

$$\dfrac{1+\sin x}{1-\sin x}=\dfrac{(1+\sin x)^2}{1-\sin^{2}x}=\dfrac{(1+\sin x)^2}{\cos^{2}x}.$$

Since $$x\in(0,\pi/2)$$, both $$1+\sin x>0$$ and $$\cos x>0,$$ hence the square root equals the positive quotient:

$$u(x)=\dfrac{1+\sin x}{\cos x}.$$

So the given function can be rewritten without the radical as

$$y=f(x)=\tan^{-1}\!\left(\dfrac{1+\sin x}{\cos x}\right).$$

To find the slope of the tangent we differentiate. The derivative of $$\tan^{-1}(\text{something})$$ is, in general,

$$\dfrac{d}{dx}\bigl[\tan^{-1}(z)\bigr]=\dfrac{1}{1+z^{2}}\dfrac{dz}{dx}.$$

Here $$z=u(x)=\dfrac{1+\sin x}{\cos x}.$$ We now compute $$u'(x).$$ Writing $$u(x)$$ as a quotient and using the quotient rule,

$$u(x)=\dfrac{1+\sin x}{\cos x}.$$

The quotient rule states that for $$\dfrac{p(x)}{q(x)}$$ we have $$\dfrac{d}{dx}\left(\dfrac{p}{q}\right)=\dfrac{q\,p'-p\,q'}{q^{2}}.$$ Here

$$p(x)=1+\sin x\quad\Rightarrow\quad p'(x)=\cos x,$$

$$q(x)=\cos x\quad\Rightarrow\quad q'(x)=-\sin x.$$

Substituting into the quotient rule gives

$$u'(x)=\dfrac{\cos x\cdot\cos x-(1+\sin x)(-\sin x)}{\cos^{2}x}.$$

Expanding the numerator carefully,

$$u'(x)=\dfrac{\cos^{2}x+(1+\sin x)\sin x}{\cos^{2}x}.$$

Next we expand the product in the numerator:

$$\cos^{2}x+(1+\sin x)\sin x=\cos^{2}x+\sin x+\sin^{2}x.$$

Using the Pythagorean identity $$\sin^{2}x+\cos^{2}x=1,$$ the numerator simplifies to

$$1+\sin x.$$

Hence

$$u'(x)=\dfrac{1+\sin x}{\cos^{2}x}.$$

Now we calculate the derivative of $$f(x).$$ Using the formula quoted earlier,

$$f'(x)=\dfrac{1}{1+u^{2}(x)}\;u'(x).$$

We already have $$u'(x).$$ We still need $$1+u^{2}(x).$$ Because

$$u(x)=\dfrac{1+\sin x}{\cos x},$$

we get

$$u^{2}(x)=\left(\dfrac{1+\sin x}{\cos x}\right)^{2}=\dfrac{(1+\sin x)^{2}}{\cos^{2}x}.$$

Therefore

$$1+u^{2}(x)=1+\dfrac{(1+\sin x)^{2}}{\cos^{2}x}=\dfrac{\cos^{2}x+(1+\sin x)^{2}}{\cos^{2}x}.$$

Putting everything together,

$$f'(x)=\dfrac{\dfrac{1+\sin x}{\cos^{2}x}}{\dfrac{\cos^{2}x+(1+\sin x)^{2}}{\cos^{2}x}}=\dfrac{1+\sin x}{\cos^{2}x+\,(1+\sin x)^{2}}.$$

We now evaluate this derivative at the required point $$x=\dfrac{\pi}{6}.$$ First we list the familiar trigonometric values:

$$\sin\dfrac{\pi}{6}=\dfrac{1}{2},\qquad \cos\dfrac{\pi}{6}=\dfrac{\sqrt3}{2}.$$

Then

$$1+\sin\dfrac{\pi}{6}=1+\dfrac12=\dfrac32,$$

$$\cos^{2}\dfrac{\pi}{6}=\left(\dfrac{\sqrt3}{2}\right)^{2}=\dfrac34,$$

$$(1+\sin\dfrac{\pi}{6})^{2}=\left(\dfrac32\right)^{2}=\dfrac{9}{4}.$$

Substituting these numbers into the expression for $$f'(x)$$ gives

$$f'\!\left(\dfrac{\pi}{6}\right)=\dfrac{\dfrac32}{\dfrac34+\dfrac94}=\dfrac{\dfrac32}{\dfrac{3+9}{4}}=\dfrac{\dfrac32}{\dfrac{12}{4}}=\dfrac{\dfrac32}{3}=\dfrac12.$$

Hence the slope of the tangent at $$x=\dfrac{\pi}{6}$$ is $$\dfrac12.$$

The slope of the normal is the negative reciprocal of the tangent’s slope. Therefore

$$m_{\text{normal}}=-2.$$

Next we need the coordinates of the point on the curve where the normal is drawn. The abscissa is already known: $$x_{0}=\dfrac{\pi}{6}.$$ The ordinate is

$$y_{0}=f\!\left(\dfrac{\pi}{6}\right)=\tan^{-1}\!\left(\dfrac{1+\sin\dfrac{\pi}{6}}{\cos\dfrac{\pi}{6}}\right)=\tan^{-1}\!\left(\dfrac{\frac32}{\frac{\sqrt3}{2}}\right)=\tan^{-1}\!\left(\dfrac{3}{\sqrt3}\right)=\tan^{-1}(\sqrt3).$$

Because $$\tan\left(\dfrac{\pi}{3}\right)=\sqrt3,$$ we have

$$y_{0}=\dfrac{\pi}{3}.$$

Thus the point of contact is

$$(x_{0},y_{0})=\left(\dfrac{\pi}{6},\dfrac{\pi}{3}\right).$$

With slope $$-2$$ and passing through that point, the equation of the normal line is obtained from the two-point form $$y-y_{0}=m(x-x_{0}).$$ Substituting, we write

$$y-\dfrac{\pi}{3}=-2\left(x-\dfrac{\pi}{6}\right).$$

To determine which given option lies on this line we check each point.

Option A: $$\left(\dfrac{\pi}{6},0\right).$$ Substituting $$x=\dfrac{\pi}{6}$$ into the right-hand side gives $$-2\!\left(\dfrac{\pi}{6}-\dfrac{\pi}{6}\right)+\dfrac{\pi}{3}=\dfrac{\pi}{3}\neq0,$$ so Option A does not lie on the normal.

Option B: $$\left(\dfrac{\pi}{4},0\right).$$ Substitution yields

$$-2\!\left(\dfrac{\pi}{4}-\dfrac{\pi}{6}\right)+\dfrac{\pi}{3}=-2\!\left(\dfrac{3\pi-2\pi}{12}\right)+\dfrac{\pi}{3}=-2\!\left(\dfrac{\pi}{12}\right)+\dfrac{\pi}{3}=-\dfrac{\pi}{6}+\dfrac{\pi}{3}=\dfrac{\pi}{6}\neq0,$$

hence Option B is rejected.

Option C: $$(0,0).$$ For $$x=0$$ the right-hand side gives $$-2\!\left(0-\dfrac{\pi}{6}\right)+\dfrac{\pi}{3}=2\!\left(\dfrac{\pi}{6}\right)+\dfrac{\pi}{3}=\dfrac{\pi}{3}+\dfrac{\pi}{3}=\dfrac{2\pi}{3}\neq0,$$ so Option C is also not on the line.

Option D: $$\left(0,\dfrac{2\pi}{3}\right).$$ Substituting $$x=0$$ into the right-hand side gives exactly what we found above, namely $$\dfrac{2\pi}{3}.$$ Therefore the left-hand side $$y$$ equals the right-hand side, so this point lies on the normal.

Hence, the correct answer is Option D.