Question

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

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

Answer 1

Answer: (A)

Eor statement p $\theta =240^{\circ }$
$2\sin \left(\frac{240^{\circ }}{2}\right)=\sqrt{1+\sin 240^{\circ }}-\sqrt{1-\sin 240^{\circ }}$
$2\sin 120^{\circ }=\sqrt{1-\frac{\sqrt{3}}{2}}-\sqrt{1+\frac{\sqrt{3}}{2}}$
$2\cdot \frac{\sqrt{3}}{2}=\sqrt{\frac{4-2\sqrt{3}}{4}}-\sqrt{\frac{4+2\sqrt{3}}{4}}$
$\sqrt{3}=\sqrt{\frac{(\sqrt{3}-1{)}^2}{4}}-\sqrt{\frac{(\sqrt{3}+1{)}^2}{4}}$
$\sqrt{3}=\frac{\sqrt{3}-1}{2}-\frac{\sqrt{3}+1}{2}$
$\sqrt{3}\ne -1$
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.