Problem 53 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

. Consider the following two statements. Statement \(p:\) The value of \(\sin 120^{\circ}\) can be divided by taking \(\theta=240^{\circ}\) in the equation \(2 \sin \frac{\theta}{2}=\sqrt{1+\sin \theta}-\sqrt{1-\sin \theta}\). Statement \(q:\) Theangles \(A, B, C\) and \(D\) of any quadrilateral \(A B C D\) 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) \(\mathrm{F}, \mathrm{T}\) (b) \(\mathrm{T}, \mathrm{T}\) (c) \(\mathrm{F}_{3} \mathrm{~F}\) (d) \(\mathrm{T}, \mathrm{F}\)

Step-by-Step Solution

Answer

The truth values are: (a) F, T.

Step 1: Understanding Statement p

The equation to verify in Statement p is \( 2 \sin \frac{\theta}{2} = \sqrt{1 + \sin \theta} - \sqrt{1 - \sin \theta} \). Given that \( \theta = 240^{\circ} \), we need to evaluate this equation.First, we find \( \sin 120^{\circ} \). Using the identity \( \sin(180^{\circ} - x) = \sin x \), we get \( \sin 120^{\circ} = \sin 60^{\circ} = \frac{\sqrt{3}}{2} \).

Step 2: Evaluating the LHS for Statement p

We calculate the left-hand side (LHS) of the equation which is \( 2 \sin \frac{\theta}{2} \) with \( \theta = 240^{\circ} \).\[\sin \frac{240^{\circ}}{2} = \sin 120^{\circ} = \frac{\sqrt{3}}{2}\]Thus, the LHS becomes \( 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} \).

Step 3: Evaluating the RHS for Statement p

Next, calculate the right-hand side (RHS) with \( \theta = 240^{\circ} \).\[\\sin 240^{\circ} = -\sin 60^{\circ} = -\frac{\sqrt{3}}{2} \\sqrt{1 + \sin 240^{\circ}} - \sqrt{1 - \sin 240^{\circ}} = \sqrt{1 - \frac{\sqrt{3}}{2}} - \sqrt{1 + \frac{\sqrt{3}}{2}} \\approx \sqrt{\frac{2 - \sqrt{3}}{2}} - \sqrt{\frac{2 + \sqrt{3}}{2}} eq \sqrt{3}\]This calculation shows that the expression is not equal to \( \sqrt{3} \). Therefore, Statement p is false.

Step 4: Understanding Statement q

For Statement q, we're given that the angles of any quadrilateral satisfy:\[\cos \left(\frac{1}{2}(A+C)\right) + \cos \left(\frac{1}{2}(B+D)\right) = 0\] Where \( A, B, C, \) and \( D \) are the angles of a quadrilateral, and \( A + B + C + D = 360^{\circ} \).

Step 5: Evaluating Statement q

Substituting \( B+D = 180^{\circ} - (A+C) \) into the equation,\[\cos \left(\frac{1}{2}(A+C)\right) + \cos \left(\frac{1}{2}(180^{\circ} - (A+C))\right) = \cos \left(\frac{1}{2}(180^{\circ})\right) = 0\]Thus, this simplifies to \( \cos(90^{\circ}) = 0 \).Hence, Statement q is true.

Key Concepts

Sine and Cosine Laws

The Sine Law and Cosine Law are fundamental tools in trigonometry. They help in solving triangles and finding unknown sides or angles. Let's explore the Sine Law first:

The Sine Law relates the sides of a triangle to the sines of its angles. It states that for any triangle \(ABC\):

\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]

where \(a, b,\) and \(c\) are the sides opposite angles \(A, B,\) and \(C\), respectively. This law is useful for non-right triangles.

The Cosine Law, on the other hand, applies to any triangle and provides a way to calculate an unknown side when two sides and the included angle are known, or an unknown angle when all sides are known. Its formula is:

\[ c^2 = a^2 + b^2 - 2ab\cos C \]

This is analogous to the Pythagorean theorem, but it applies to all triangles, not just right-angled ones. Both laws play a crucial role in understanding complex trigonometric identities and equations, such as those involved in Statement \(p\) from the exercise.

Quadrilateral Properties

Quadrilaterals are four-sided polygons that have a variety of types, such as rectangles, squares, trapezoids, and parallelograms. Each possesses unique properties. However, there are some universal properties true for all quadrilaterals:

• The sum of interior angles is always \(360^{\circ}\).
• They have four sides, four angles, and four vertices.

In the context of Statement \(q\), it's key to understand how the sum of angles plays into the given trigonometric condition. The equation \(\cos \left(\frac{1}{2}(A+C)\right) + \cos \left(\frac{1}{2}(B+D)\right) = 0\) is a curious condition that evaluates the relationship between angles of a quadrilateral. This specific equation arises from the balance of certain symmetrical properties, similar to those seen in cyclic quadrilaterals where opposite angles sum to \(180^{\circ}\). Understanding these properties helps in verifying geometrical propositions and solving such exercises effectively.

Angles in Geometry

Angles are a core part of geometry. They are everywhere — from triangles and quadrilaterals to polygons with more sides. Here are some fundamentals about angles to solidify understanding:

• The angle sum of a triangle is always \(180^{\circ}\).
• The angle sum of any polygon can be calculated by \((n-2) \times 180^{\circ}\), where \(n\) is the number of sides.

In quadrilaterals specifically, knowing that the sum of the angles is \(360^{\circ}\) forms the basis for many geometric derivations and proofs, just like in our Statement \(q\). Also, when dealing with trigonometric functions, relationships like those in the statement help simplify and solve for unknown angles inside different-shaped quadrilaterals.

Understanding angles' interactions helps us make sense of why certain expressions, like \(\cos\left(\frac{1}{2}(A+C)\right) + \cos\left(\frac{1}{2}(B+D)\right) = 0\), come to fruition in geometric proofs. This foundation allows us to apply and verify trigonometric identities correctly.