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

1. Let's evaluate the statement: "The value of $\sin \, 120^{\circ}$ can be derived by taking $\theta = 240^{\circ}$ in the equation $2 \sin \, \frac{\theta}{2} = \sqrt{1 + \sin \, \theta} - \sqrt{1 - \sin \, \theta}$."
   • First, understand the given equation: $2 \sin \, \frac{\theta}{2} = \sqrt{1 + \sin \, \theta} - \sqrt{1 - \sin \, \theta}$.
   • Now, if we substitute $\theta = 240^{\circ}$, the left-hand side becomes $2 \sin \, \frac{240}{2} = 2 \sin 120^{\circ}$.
   • Since $\sin 120^{\circ} = \sin (180^{\circ} - 60^{\circ}) = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$, the left-hand side becomes $2 \times \frac{\sqrt{3}}{2} = \sqrt{3}$.
   • Now evaluate the right-hand side:
     • $\sin 240^{\circ} = \sin (180^{\circ} + 60^{\circ}) = -\sin 60^{\circ} = -\frac{\sqrt{3}}{2}$.
     • So, the right-hand side becomes: $\sqrt{1 - \frac{\sqrt{3}}{2}} - \sqrt{1 + \frac{\sqrt{3}}{2}}$.
   • This value does not simplify to $\sqrt{3}$, hence the equation does not hold.
   • Thus, the first statement is false.
2. Now, consider the second statement: "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 $."
   • In any quadrilateral, the sum of its interior angles is $360^{\circ}$, i.e., $A + B + C + D = 360^{\circ}$.
   • Thus, we have $A + C = 360^{\circ} - (B + D)$.
   • Therefore, $\frac{1}{2}(A+C) = \frac{1}{2}(360^{\circ} - (B+D))=180^{\circ}- \frac{1}{2}(B+D)$.
   • Using the trigonometric identity, $\cos(180^{\circ} - x) = -\cos(x)$, we get: $\cos\left(\frac{1}{2}(A+C)\right) = -\cos\left(\frac{1}{2}(B+D)\right)$.
   • Thus, $\cos\left(\frac{1}{2}(A+C)\right) + \cos\left(\frac{1}{2}(B+D)\right) = 0$, hence the second statement is true.

Based on the evaluations above, the truth values of the statements are "False" for the first and "True" for the second. Therefore, the correct answer is: F, T.