Problem 8 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

In triangle \(A B C\), prove that, \(\frac{\text { area of the in-circle }}{\text { area of triangle } A B C}\) \(=\frac{\pi}{\cot \left(\frac{A}{2}\right) \cdot \cot \left(\frac{B}{2}\right) \cdot \cot \left(\frac{C}{2}\right)}\)

Step-by-Step Solution

Answer

The proof shows that the provided equality indeed holds. The key to the solution is to make use of the formula linking the area of a triangle to the inradius and semiperimeter, and to use trigonometric half-angle identities. Both sides of the equation end up being equal to \(\frac{\pi r}{s} \), affirming that the equation is correct.

Step 1: Assigning Variables

First, let's denote the sides of the triangle \(ABC\) as \(a, b, c\), where \(a = BC\), \(b = AC\), and \(c = AB\), and the semiperimeter of the triangle as \(s = \frac{a+b+c}{2}\). The area of the triangle is then given by the formula: \(K = rs\), where \(r\) is the radius of the incircle.

Step 2: Trigonometric Identities

Next, recall the trigonometric identities for the half-angles: \(\cot \frac{A}{2} = \frac{1+\cos A}{\sin A} = \frac{s-b}{r}\), \(\cot \frac{B}{2} = \frac{s-c}{r}\), and \(\cot \frac{C}{2} = \frac{s-a}{r}\). Also, \(\sin A = \frac{2K}{bc}\).

Step 3: Substituting Values

Now substitute the trigonometric identities into the formula provided in the exercise and simplify: \(\frac{\pi}{\cot \frac{A}{2} \cdot \cot \frac{B}{2} \cdot \cot \frac{C}{2}} = \frac{\pi r^3}{(s-a)\cdot (s-b) \cdot (s-c)}\)

Step 4: Proving the Identity

Finally, note that the area of incircle equals \(\pi r^2\). So, the left-hand side of the equation to be proved is the ratio of the area of incircle to the area of triangle, which equals \(\frac{\pi r^2}{rs} = \frac{\pi r}{s} = \frac{\pi r^3}{(s-a)\cdot (s-b) \cdot (s-c)}\), which matches the right-hand side of the equation obtained in step 3.

Key Concepts

Area of Triangle

Finding the area of a triangle is one of the essential skills in geometry. There are several ways to calculate it, but one of the most useful formulas involves the incircle, which is the circle snugly fitting inside the triangle, touching all three sides.
In this context, the area of the triangle is calculated using the formula:

• The area, denoted as \(K\), is equal to \(rs\), where \(r\) is the radius of the incircle and \(s\) is the semi-perimeter of the triangle.
• The semi-perimeter \(s\) can be found by halving the sum of all sides: \(s = \frac{a+b+c}{2}\).

Understanding the relationship between the side lengths of the triangle and its area is crucial, especially when proving identities involving the incircle.

Half-Angle Identities

Half-angle identities are trigonometric formulas that relate the half-angle values of sine, cosine, and cotangent functions to the values of other trigonometric functions. These identities are particularly useful in the context of incircles.
For a triangle, they help to simplify expressions involving angles, such as:

• \( \cot \frac{A}{2} = \frac{1+\cos A}{\sin A} = \frac{s-b}{r} \)
• \( \cot \frac{B}{2} = \frac{s-c}{r} \)
• \( \cot \frac{C}{2} = \frac{s-a}{r} \)

Using half-angle identities makes complex expressions more manageable, particularly when you need to manipulate and equate trigonometric expressions within geometric contexts.

Trigonometric Identities

Trigonometric identities are equations that involve trigonometric functions and are true for all values of the involved variables. They facilitate the expression simplification process in triangles.
In the context of triangle \(ABC\):

• It is crucial to understand that \( \sin A = \frac{2K}{bc} \), where \(K\) is the area expressed not only in terms of incircle radius.
• This identity links the angle \(A\) with the relationships between the triangle's sides and area.

Trigonometric identities are inherent tools used to simplify calculations involving angles and sides, making them indispensable in geometric proofs like the one in the original exercise.

Incircle Radius

The incircle of a triangle is the largest possible circle that fits snugly inside and touches each side of the triangle exactly once. The radius of this incircle holds a significant role in not only calculating the triangle's area but also in proving related identities.
The radius \(r\) can be found using the following relationship:

• \(r = \frac{K}{s}\), where \(K\) is the area and \(s\) is the semi-perimeter of the triangle.
• This makes the radius a bridge linking the area and perimeter, which is why it's so effective in solving various trigonometric problems concerning triangles.

Understanding how the incircle radius interacts with other components of the triangle allows for a deeper comprehension of such geometric proofs.