Question

Consider the following two statements : The value of $\sin \, 120^{\circ}$ can be derived by taking $\theta =240^{\circ}$ in the equation $2 \sin \, \frac{\theta}{2} = \sqrt{1 + \sin \, \theta} - \sqrt{1 - \sin \, \theta}$. 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 :

• A. F, T
• B. T, F
• C. T, T
• D. F, F

Answer 1

The Correct Option is A

Eor statement p $\theta=240^{\circ}$
$2 \sin \left(\frac{240^{\circ}}{2}\right) =\sqrt{1+\sin 240^{\circ}}-\sqrt{1-\sin 240^{\circ}} $
$2 \sin 120^{\circ} =\sqrt{1-\frac{\sqrt{3}}{2}}-\sqrt{1+\frac{\sqrt{3}}{2}} $
$2 \cdot \frac{\sqrt{3}}{2} =\sqrt{\frac{4-2 \sqrt{3}}{4}}-\sqrt{\frac{4+2 \sqrt{3}}{4}} $
$\sqrt{3} =\sqrt{\frac{(\sqrt{3}-1)^{2}}{4}}-\sqrt{\frac{(\sqrt{3}+1)^{2}}{4}}$
$\sqrt{3} =\frac{\sqrt{3}-1}{2}-\frac{\sqrt{3}+1}{2} $
$\sqrt{3} \neq-1$
Therefore, statement $p$ is false.
For statement q: $\cos \left(\frac{1}{2}(A+C)\right)+\cos \left(\frac{1}{2}(B+D)\right)=0$
$A+B+C+D=2 \pi$
$ \Rightarrow \frac{A+C}{2}=\pi-\left(\frac{B+D}{2}\right) $
$\cos \frac{A+C}{2}+\cos \frac{B+D}{2} \cos \left(\pi-\frac{B+D}{2}\right)+\cos \left(\frac{B+D}{2}\right)=0$
Therefore, statement $q$ is true.