Statement p Analysis

We need to verify if $2\sin \frac{\theta}{2} = \sqrt{1+\sin \theta} - \sqrt{1-\sin \theta}$ holds true for $\theta = 240^\circ$.

Let $\theta = 240^\circ$, then $\frac{\theta}{2} = 120^\circ$. The left side (LHS) is $2\sin(120^\circ) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$.

For the right side (RHS), we use the identities: $1 + \sin \theta = (\sin(\theta/2) + \cos(\theta/2))^2$ $1 - \sin \theta = (\cos(\theta/2) - \sin(\theta/2))^2$ So, $\sqrt{1+\sin \theta} = |\sin(\theta/2) + \cos(\theta/2)|$ And $\sqrt{1-\sin \theta} = |\cos(\theta/2) - \sin(\theta/2)|$

Substitute $\theta/2 = 120^\circ$: $\sin(120^\circ) = \frac{\sqrt{3}}{2}$ $\cos(120^\circ) = -\frac{1}{2}$

Calculate the terms inside the absolute values: $\sin(120^\circ) + \cos(120^\circ) = \frac{\sqrt{3}}{2} - \frac{1}{2} = \frac{\sqrt{3}-1}{2}$ (Positive) $\cos(120^\circ) - \sin(120^\circ) = -\frac{1}{2} - \frac{\sqrt{3}}{2} = -\frac{1+\sqrt{3}}{2}$ (Negative)

Evaluate the absolute values: $|\sin(120^\circ) + \cos(120^\circ)| = \frac{\sqrt{3}-1}{2}$ $|\cos(120^\circ) - \sin(120^\circ)| = |-\frac{1+\sqrt{3}}{2}| = \frac{1+\sqrt{3}}{2}$

RHS = $|\sin(120^\circ) + \cos(120^\circ)| - |\cos(120^\circ) - \sin(120^\circ)| = \frac{\sqrt{3}-1}{2} - \frac{1+\sqrt{3}}{2} = \frac{\sqrt{3}-1 - 1 - \sqrt{3}}{2} = \frac{-2}{2} = -1$.

Since LHS ($\sqrt{3}$) is not equal to RHS (-1), Statement p is False.

Statement q Analysis

Let A, B, C, D be the angles of a quadrilateral. The sum of the angles is $A+B+C+D = 360^\circ$. We need to check if $\cos \left( \frac{1}{2}(A+C) \right) + \cos \left( \frac{1}{2}(B+D) \right) = 0$.

From the sum of angles, $A+C = 360^\circ - (B+D)$. Dividing by 2, we get $\frac{1}{2}(A+C) = \frac{1}{2}(360^\circ - (B+D)) = 180^\circ - \frac{1}{2}(B+D)$.

Substitute this into the equation: $\cos \left( 180^\circ - \frac{1}{2}(B+D) \right) + \cos \left( \frac{1}{2}(B+D) \right)$

Using the trigonometric identity $\cos(180^\circ - x) = -\cos x$, where $x = \frac{1}{2}(B+D)$: $-\cos \left( \frac{1}{2}(B+D) \right) + \cos \left( \frac{1}{2}(B+D) \right) = 0$.

The equation $0 = 0$ is true. Therefore, Statement q is True.

Conclusion

Statement p is False (F) and Statement q is True (T). The truth values are (F, T).