Consider the following two statements.
Statement p: The value of $$\sin 120^\circ$$ can be divided by taking $$\theta = 240^\circ$$ in the equation $$2\sin\frac{\theta}{2} = \sqrt{1 + \sin\theta} - \sqrt{1 - \sin\theta}$$.
Statement q: The angles A, B, C and D of any quadrilateral ABCD satisfy the equation $$\cos\left(\frac{1}{2}(A+C)\right) + \cos\left(\frac{1}{2}(B+D)\right) = 0$$.
Then the truth values of p and q are respectively:

First we examine Statement p. The equation given is

$$2\sin\frac{\theta}{2}= \sqrt{\,1+\sin\theta\,}-\sqrt{\,1-\sin\theta\,}.$$

We are told to put $$\theta =240^\circ.$$ Let us evaluate every term one by one.

We have $$\sin 240^\circ=\sin\left(180^\circ+60^\circ\right)=-\sin60^\circ=-\frac{\sqrt3}{2}.$$

Now compute the left-hand side (L.H.S.) of the equation:

$$2\sin\frac{\theta}{2}=2\sin\left(\frac{240^\circ}{2}\right)=2\sin120^\circ.$$

Using $$\sin120^\circ=\sin(60^\circ)=\frac{\sqrt3}{2},$$ we get

$$2\sin120^\circ=2\cdot\frac{\sqrt3}{2}=\sqrt3.$$

So the L.H.S. equals $$\sqrt3\;(\approx1.732).$$

Next we evaluate the right-hand side (R.H.S.):

$$\sqrt{\,1+\sin\theta\,}-\sqrt{\,1-\sin\theta\,}.$$

Substituting $$\sin\theta=-\frac{\sqrt3}{2},$$ we obtain

$$\sqrt{\,1-\frac{\sqrt3}{2}\,}-\sqrt{\,1+\frac{\sqrt3}{2}\,}.$$

Convert the 1 to a fraction with denominator 2:

$$1=\frac{2}{2},$$ so

$$\sqrt{\frac{2-\sqrt3}{2}}-\sqrt{\frac{2+\sqrt3}{2}}.$$

The first square root is approximately $$\sqrt{0.134}\approx0.366,$$ and the second is approximately $$\sqrt{1.866}\approx1.366.$$ Therefore the R.H.S. is roughly

$$0.366-1.366\approx-1.000.$$

We now compare:

$$\text{L.H.S.}=\sqrt3\;(\approx1.732),\qquad\text{R.H.S.}\approx-1.000.$$

The two sides are not equal, so the equation does not hold for $$\theta=240^\circ.$$ Hence Statement p is false.

Now we investigate Statement q. In any quadrilateral ABCD the interior angles satisfy

$$A+B+C+D=360^\circ.$$

We are to test the identity

$$\cos\!\left(\tfrac12(A+C)\right)+\cos\!\left(\tfrac12(B+D)\right)=0.$$

Let us denote

$$X=\tfrac12(A+C).$$

Because the sum of all four angles is $$360^\circ,$$ we have

$$B+D=360^\circ-(A+C).$$

Dividing by 2 gives

$$\tfrac12(B+D)=\tfrac12\!\left[\,360^\circ-(A+C)\right]=180^\circ-\tfrac12(A+C)=180^\circ-X.$$

Now write the left side of the proposed identity in terms of $$X:$$

$$\cos X+\cos(180^\circ-X).$$

We recall the trigonometric fact: $$\cos(180^\circ-\theta)=-\cos\theta.$$ Stating the formula plainly, “cosine changes sign in the second quadrant.” Using this formula with $$\theta=X$$ gives

$$\cos(180^\circ-X)=-\cos X.$$

Substituting, we get

$$\cos X+\bigl(-\cos X\bigr)=0.$$

The expression indeed equals zero for every choice of angles A, B, C and D that add to $$360^\circ.$$ Thus Statement q is true.

We have found:

Statement p: False,   Statement q: True.

The ordered pair of truth values is (F, T), which corresponds to Option A.

Hence, the correct answer is Option A.