Statement - I : The value of the integral $$\int_{\pi/6}^{\pi/3} \frac{dx}{1 + \sqrt{\tan x}}$$ is equal to $$\frac{\pi}{6}$$.
Statement - II : $$\int_a^b f(x)dx = \int_a^b f(a + b - x)dx$$.

We begin with Statement - II because the property it claims can be proved in a few elementary steps and will later be useful.

We want to show that for any continuous function $$f(x)$$ and any real numbers $$a$$ and $$b$$ (with $$a<b$$) we have the identity

$$\int_{a}^{b} f(x)\,dx \;=\; \int_{a}^{b} f(a+b-x)\,dx.$$

To verify it, we perform the substitution $$x=a+b-t.$$ First we write down the relations that follow from this change of variable:

When $$x=a+b-t,$$ then $$t=a+b-x,$$ and hence $$dt=-dx.$$ The limits also transform:
• When $$x=a,$$ we get $$t=a+b-a=b.$$
• When $$x=b,$$ we get $$t=a+b-b=a.$$
So the integral becomes

$$ \int_{x=a}^{x=b} f(x)\,dx \;=\;\int_{t=b}^{t=a} f(a+b-t)\,(-dt) \;=\;\int_{t=a}^{t=b} f(a+b-t)\,dt. $$

The dummy variable $$t$$ may be renamed $$x$$ without altering the value of the integral, giving exactly

$$\int_{a}^{b} f(x)\,dx=\int_{a}^{b} f(a+b-x)\,dx,$$

which is what Statement - II asserts. Thus, Statement - II is true.

Next we turn to Statement - I and examine the definite integral

$$I=\int_{\pi/6}^{\pi/3}\frac{dx}{\,1+\sqrt{\tan x}\,}.$$

The length of the interval of integration is easily computed:

$$b-a=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}.$$

Inside the integral we have the denominator $$1+\sqrt{\tan x}.$$ On $$\left[\dfrac{\pi}{6},\dfrac{\pi}{3}\right]$$ the tangent is positive, so $$\sqrt{\tan x}>0.$$ Therefore

$$1+\sqrt{\tan x}\;>\;1\quad\Rightarrow\quad \frac{1}{1+\sqrt{\tan x}}\;<\;1.$$

Because the integrand is strictly less than 1 at every point of the interval, the entire integral must be strictly less than the length of the interval. Writing this inequality step by step, we have

$$ \forall x\in\left[\frac{\pi}{6},\frac{\pi}{3}\right]: \;0<\frac{1}{1+\sqrt{\tan x}}<1 $$ and hence $$ 0<\int_{\pi/6}^{\pi/3}\frac{dx}{1+\sqrt{\tan x}} \;<\;\int_{\pi/6}^{\pi/3}1\,dx \;=\;\frac{\pi}{3}-\frac{\pi}{6} \;=\;\frac{\pi}{6}. $$

Thus the integral $$I$$ is strictly smaller than $$\dfrac{\pi}{6}.$$ It can never be equal to $$\dfrac{\pi}{6},$$ as claimed by Statement - I. Consequently, Statement - I is false.

Summarising our findings:

• Statement - I is false.
• Statement - II is true.

The option that matches this combination is Option B (the second option in the list).

Hence, the correct answer is Option B.