Question

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

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

Answer 1

The correct option is D $F,T$

$Statementp:$
$2\sin \frac{\theta }{2}=\sqrt{1+\sin \theta }-\sqrt{1-\sin \theta }$
put $\theta ={240}^{\circ }$
$2\sin \frac{{240}^{\circ }}{2}=\sqrt{1+\sin {240}^{\circ }}-\sqrt{1-\sin {240}^{\circ }}$
$2\sin {120}^{\circ }=\sqrt{1-\frac{\sqrt{3}}{2}}-\sqrt{1+\frac{\sqrt{3}}{2}}$
which is not equal
so statement p is false.
$Statementq:$
$\cos \left(\frac{1}{2}\right(A+C\left)\right)+\cos \left(\frac{1}{2}\right(B+D\left)\right)=0$
From L.H.S
$\cos \left(\frac{1}{2}\right(A+C\left)\right)+\cos \left(\frac{1}{2}\right(B+D\left)\right)$
$=2\cos \left(\frac{\frac{A+C+B+D}{2}}{2}\right)\cdot \cos \left(\frac{\frac{A+C-B-D}{2}}{2}\right)$
As, $A+B+C+D={360}^{\circ }$
$=2\cos \left(\frac{{360}^{\circ }}{4}\right)\cdot \cos \left(\frac{\frac{A+C-B-D}{2}}{2}\right)$
$=0$
Hence statement q is true.