Consider the following two statements :- Statement p : The value of sin 120° can be derived by takin

Q: Wait, why did the RHS of Statement p result in -1 instead of $\sqrt{3}$? Did I mess up the square root?

You likely forgot the absolute value property! Remember that $\sqrt{1+\sin \theta} = |\sin \frac{\theta}{2} + \cos \frac{\theta}{2}|$. At $120^\circ$, the cosine term is negative and larger in magnitude than the sine term, which flips the sign inside the absolute value. Always check the quadrant!

Q: Why can't we just say $\sqrt{1+\sin \theta} = \sin \frac{\theta}{2} + \cos \frac{\theta}{2}$ without the absolute value bars?

Because the square root function, by definition, returns a non-negative value. If the expression inside the square evaluates to a negative number, the identity $\sqrt{x^2} = x$ fails. We must use $|x|$ to ensure the result remains positive.

Q: For Statement q, does this only work for cyclic quadrilaterals?

Great question! Actually, no. The property $A+B+C+D = 360^\circ$ holds for *any* convex quadrilateral. Since we only used the sum of the angles, the result is universal for all quadrilaterals, not just cyclic ones.

Q: Is there a shortcut to check Statement p without calculating the full values?

Yes! Look at the signs. At $\theta = 240^\circ$, the LHS is $2 \sin 120^\circ$, which is positive. However, the RHS involves the difference of two square roots. By checking the magnitude of the terms, you can quickly see the RHS will be negative, making equality impossible.

Q: Why does $\cos(\frac{A+C}{2}) + \cos(\frac{B+D}{2}) = 0$ always hold?

It holds because $\frac{A+C}{2}$ and $\frac{B+D}{2}$ are supplementary angles (they add up to $180^\circ$). Since $\cos(180^\circ - \theta) = -\cos \theta$, the two cosine terms will always cancel each other out perfectly.

Question

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

Options:

(A) T, T

(B) F, F

(C) F, T

(D) T, F

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we aren't just solving a problem; we are dissecting the delicate anatomy of trigonometric identities.

Let us embark on this journey to evaluate the truth values of two statements, $p$ and $q$.

### Statement $p$: The Trap of the Square Root

Imagine you are standing at $	heta = 240^\circ$. The statement claims that $2 \sin \frac{	heta}{2} = \sqrt{1+\sin 	heta} - \sqrt{1-\sin 	heta}$.

First, let us calculate the Left Hand Side (LHS). Substituting $	heta = 240^\circ$, we get $2 \sin(120^\circ)$.

Since $\sin 120^\circ = \frac{\sqrt{3}}{2}$, our LHS simplifies to:

$$LHS = 2 \left( \frac{\sqrt{3}}{2} ight) = \sqrt{3}$$

Now, consider the Right Hand Side (RHS): $\sqrt{1+\sin 240^\circ} - \sqrt{1-\sin 240^\circ}$. Recall that $\sin 240^\circ = -\frac{\sqrt{3}}{2}$.

The identity $\sqrt{1 \pm \sin 	heta} = |\sin \frac{	heta}{2} \pm \cos \frac{	heta}{2}|$ is our guiding light. When we plug in $	heta = 240^\circ$, we evaluate the expression at $\frac{	heta}{2} = 120^\circ$.

Note that $\sin 120^\circ = \frac{\sqrt{3}}{2}$ and $\cos 120^\circ = -\frac{1}{2}$. The expression becomes:

$$RHS = \left| \sin 120^\circ + \cos 120^\circ ight| - \left| \sin 120^\circ - \cos 120^\circ ight|$$

$$RHS = \left| \frac{\sqrt{3}}{2} - \frac{1}{2} ight| - \left| \frac{\sqrt{3}}{2} + \frac{1}{2} ight|$$

Since $\frac{\sqrt{3}}{2} > \frac{1}{2}$, the first absolute value is positive, and the second is also positive. Thus:

$$RHS = \left( \frac{\sqrt{3}-1}{2} ight) - \left( \frac{\sqrt{3}+1}{2} ight) = -1$$

Since $\sqrt{3} 
eq -1$, Statement $p$ is undeniably **False**. Always respect the quadrant and the absolute value signs!

### Statement $q$: The Universal Harmony of Quadrilaterals

Now, let us shift our gaze to the geometry of a quadrilateral $ABCD$. We are asked if $\cos \left( \frac{A+C}{2} ight) + \cos \left( \frac{B+D}{2} ight) = 0$.

The sum of the interior angles of a quadrilateral is $360^\circ$. That is, $A + B + C + D = 360^\circ$.

Dividing this equation by $2$, we obtain:

$$\frac{A+C}{2} + \frac{B+D}{2} = 180^\circ$$

Let $X = \frac{A+C}{2}$ and $Y = \frac{B+D}{2}$. We have discovered that $X + Y = 180^\circ$, which implies $Y = 180^\circ - X$.

Now, substitute this into the expression $\cos X + \cos Y$:

$$\cos X + \cos(180^\circ - X) = \cos X - \cos X = 0$$

The result is $0$. It cancels out perfectly, confirming that Statement $q$ is **True**.

### The Final Verdict

Through our investigation, we have found that Statement $p$ is **False** and Statement $q$ is **True**.

Remember, the beauty of JEE mathematics lies not in memorizing formulas, but in understanding the constraints. Whether it is the absolute value sign hiding in a square root or the supplementary nature of angles in a quadrilateral, stay vigilant and keep pushing forward.

Select Answer:

Visualized Solution

Introduction to the Problem

Statement $p$ : Left Hand Side

Evaluating LHS

Statement $p$ : Right Hand Side

Simplifying the Square Roots

Evaluating Absolute Values

Final RHS Calculation

Conclusion for Statement $p$

Statement $q$ : Quadrilateral Angles

Rearranging the Angle Sum

Applying Cosine Identity

Conclusion for Statement $q$

Final Truth Values

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

Statement $p$

Statement $q$

The Final Verdict

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Select Answer:

Visualized Solution

Introduction to the Problem

Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p: Left Hand Side

Evaluating LHS

Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p: Right Hand Side

Simplifying the Square Roots

Evaluating Absolute Values

Final RHS Calculation

Conclusion for Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q: Quadrilateral Angles

Rearranging the Angle Sum

Applying Cosine Identity

Conclusion for Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q

Final Truth Values

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Statement .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q

The Final Verdict

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