Let PQR be a triangle of area Δ with a = 2, b = 7/2 and c = 5/2, where a, b, and c are the lengths o

Q: Why can't we just find the angle $P$ using the Cosine Rule directly?

While you *could* find $\cos P$ using the Cosine Rule, it is often computationally heavy and prone to arithmetic errors. The half-angle formula is a 'shortcut' that directly links the trigonometric ratio to the side lengths, saving you time and reducing the chance of mistakes—a crucial skill for JEE Advanced.

Q: Why are the options in terms of $\Delta$?

This is a classic JEE trap! The problem setter wants to see if you can manipulate algebraic expressions and relate them to geometric properties. Since $\Delta$ is a constant for this specific triangle, expressing the answer in terms of $\Delta$ is just a way to test your ability to substitute and verify identities.

Q: Is there any ambiguity with the sign of $\tan \frac{P}{2}$?

Great question! Since $P$ is an angle in a triangle, $0 < P < \pi$, which means $0 < P/2 < \pi/2$. In this range, $\tan(P/2)$ is always positive. Since we are dealing with $\tan^2(P/2)$, the sign is always positive, so you don't need to worry about negative values.

Question

Let $PQR$ be a triangle of area $\Delta$ with $a = 2, b = 7/2$ and $c = 5/2$, where $a, b$, and $c$ are the lengths of the sides of the triangle opposite to the angles at $P, Q$ and $R$ respectively. Then $\frac{2 \sin P - \sin 2P}{2 \sin P + \sin 2P}$ equals

Options:

(A) $\frac{3}{4\Delta}$

(B) $\frac{45}{4\Delta}$

(C) $\left(\frac{3}{4\Delta}\right)^2$

(D) $\left(\frac{45}{4\Delta}\right)^2$

Sigma Percentile · Accepted Answer

### The Symphony of Geometry and Trigonometry: A JEE Masterclass

Welcome, my dear student. Today, we are not just solving a problem; we are embarking on a journey. We are going to take a seemingly daunting trigonometric expression and watch it dissolve into elegance.

In the world of JEE Advanced, the most complex-looking problems often hide the simplest truths. Let us peel back the layers of this triangle problem together.

### Phase 1: The Algebraic Dance

We are presented with a triangle $PQR$ with sides $a=2$, $b=7/2$, and $c=5/2$. We are asked to evaluate the expression:

$$E = \frac{2 \sin P - \sin 2P}{2 \sin P + \sin 2P}$$

When you see an expression like this, your first instinct might be to panic. Don't. Look at the structure. We have $\sin P$ and $\sin 2P$.

Whenever you see a double angle, your internal alarm should ring: "Identity!" We know that $\sin 2P = 2 \sin P \cos P$. Let us substitute this into our expression:

$$E = \frac{2 \sin P - 2 \sin P \cos P}{2 \sin P + 2 \sin P \cos P}$$

Look at that! We have a common factor of $2 \sin P$ in both the numerator and the denominator. We can factor it out:

$$E = \frac{2 \sin P (1 - \cos P)}{2 \sin P (1 + \cos P)}$$

Assuming $\sin P 
eq 0$ (which is true for any triangle), we can cancel the $2 \sin P$ terms. We are left with:

$$E = \frac{1 - \cos P}{1 + \cos P}$$

This, my friend, is a classic trigonometric identity. Recall the half-angle formulas: $1 - \cos P = 2 \sin^2 \frac{P}{2}$ and $1 + \cos P = 2 \cos^2 \frac{P}{2}$.

Substituting these in, the twos cancel, and we are left with $\frac{\sin^2 \frac{P}{2}}{\cos^2 \frac{P}{2}}$, which is simply $	an^2 \frac{P}{2}$. We have tamed the beast!

### Phase 2: The Geometric Bridge

Now, we have reduced the problem to finding $	an^2 \frac{P}{2}$. But we don't have the angle $P$. We have the sides.

This is where the beauty of triangle properties shines. There is a direct, powerful bridge between the trigonometry of half-angles and the side lengths of a triangle:

$$	an^2 \frac{P}{2} = \frac{(s-b)(s-c)}{s(s-a)}$$

where $s$ is the semi-perimeter. Let us calculate $s$ first:

$$s = \frac{a+b+c}{2} = \frac{2 + 3.5 + 2.5}{2} = \frac{8}{2} = 4$$

Now, we substitute our values:

$$	an^2 \frac{P}{2} = \frac{(4 - 3.5)(4 - 2.5)}{4(4 - 2)} = \frac{0.5 	imes 1.5}{4 	imes 2} = \frac{0.75}{8} = \frac{3/4}{8} = \frac{3}{32}$$

### Phase 3: The Final Synthesis

We have our numerical answer: $3/32$. However, the options are in terms of the area $\Delta$. We must find $\Delta$ using Heron's Formula: $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$.

$$\Delta = \sqrt{4(4-2)(4-3.5)(4-2.5)} = \sqrt{4 	imes 2 	imes 0.5 	imes 1.5} = \sqrt{4 	imes 2 	imes \frac{1}{2} 	imes \frac{3}{2}} = \sqrt{6}$$

Now, let us look at the options. Option (C) is $\left(\frac{3}{4\Delta}ight)^2$. Let us test it:

$$\left(\frac{3}{4\sqrt{6}}ight)^2 = \frac{9}{16 	imes 6} = \frac{9}{96} = \frac{3}{32}$$

It matches perfectly! We have traversed the path from a complex trigonometric expression to a geometric identity, and finally to the area of the triangle.

This is the essence of JEE Advanced mathematics—connecting disparate concepts to reveal a single, elegant truth. Keep practicing, keep questioning, and most importantly, keep falling in love with the process. The final answer is **$3/32$**.

Select Answer:

Visualized Solution

Visualizing Triangle $P QR$

Simplifying the Trigonometric Expression

Substituting the Identity

Reducing to Half-Angle

Relating Angle to Sides

Calculating Semi-perimeter $s$

Substituting Values into Half-Angle Formula

Evaluating $tan^{2} \frac{P}{2}$

Finding Area $Δ$ using Heron's Formula

Calculating Area $Δ$

Matching with Options

The Sigma Insight: Properties of Triangles

The Symphony of Geometry and Trigonometry

Phase 1

Phase 2

Phase 3

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Select Answer:

Visualized Solution

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

Simplifying the Trigonometric Expression

Substituting the Identity

Reducing to Half-Angle

Relating Angle to Sides

Calculating Semi-perimeter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } s

Substituting Values into Half-Angle Formula

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan22P​

Finding Area .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Δ using Heron's Formula

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