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

Hint: First we are going to use the formula of sin2x and some other trigonometric formula and then we will check that statement p is true or not, for statement q we will use the formula for cosA + cosB and one other fact that the sum of all the angles of quadrilateral is 360.

Complete step-by-step answer:

First we are going to state all the formulas that we are going to use,
$\begin{align}
  & \cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) \\
 & {{\sin }^{2}}x+{{\cos }^{2}}x=1 \\
\end{align}$
Now we will solve the statement p,
$2\sin \dfrac{\theta }{2}=\sqrt{1+\sin \theta }-\sqrt{1-2\theta }$
$\begin{align}
  & 2\sin \dfrac{\theta }{2}=\sqrt{{{\sin }^{2}}\dfrac{\theta }{2}+{{\cos }^{2}}\dfrac{\theta }{2}+2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}-\sqrt{1-2\theta } \\
 & 2\sin \dfrac{\theta }{2}=\sqrt{{{\left( \sin \dfrac{\theta }{2}+\cos \dfrac{\theta }{2} \right)}^{2}}}-\sqrt{1-2\theta } \\
 & 2\sin \dfrac{\theta }{2}=\sin \dfrac{\theta }{2}+\cos \dfrac{\theta }{2}-\sqrt{1-2\theta } \\
 & \sin \dfrac{\theta }{2}-\cos \dfrac{\theta }{2}=-\sqrt{1-2\theta } \\
\end{align}$
Now we are going to put $\theta $ = 240 and see if it is true or not.
$\begin{align}
  & \sin 120-\cos 120 \\
 & =\dfrac{\sqrt{3}+1}{2} \\
\end{align}$
Hence we can see that the LHS is positive but the RHS is $-\sqrt{1-\dfrac{2\times 4\pi }{3}}$ which is negative.
Hence statement p is false.
Now we will look at statement q that the sum of all the four angles of the quadrilateral is 360.
A + B + C + D = 360
Now we will use this formula,
$\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$
Applying the above formula in statement q we get,
$\begin{align}
  & \cos \left( \dfrac{A+C}{2} \right)+\cos \left( \dfrac{B+D}{2} \right) \\
 & =2\cos \left( \dfrac{A+B+C+D}{4} \right)\cos \left( \dfrac{A+C-B-D}{4} \right) \\
 & =2\cos \left( \dfrac{360}{4} \right)\cos \left( \dfrac{A+C-B-D}{4} \right) \\
 & =2\cos 90\cos \left( \dfrac{A+C-B-D}{4} \right) \\
 & =0 \\
\end{align}$
As cos90 = 0.
Hence statement q is true.
Hence the correct option is (a).

Note: The formulas that we have used for solving the above question is important and the fact that we have used the sum of all the four angles of quadrilateral is 360 is also very important which helped us to solve this question.