Problem 19 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

In any triangle \(A B C\), prove that, \(4 \Delta(\cot A+\cot B+\cot C)=a^{2}+b^{2}+c^{2}\)

Step-by-Step Solution

Answer

Hence, it is proved that: \(4 \Delta(\cot A+\cot B+\cot C)=a^{2}+b^{2}+c^{2}\)

Step 1: Derive Area and Replace in Given Equation.

The area \(\Delta\) of any triangle with sides \(a\), \(b\), and \(c\), and the included angle \(\gamma\) (opposite to side \(c\)) is given by \(\Delta = \frac{1}{2}ab\sin\gamma\). Replacing \(\gamma\) with \(C\) (as it is the included angle corresponding to side \(c\)), the area can be rewritten as \(\Delta = \frac{1}{2}ab\sin C\). Now we substitute this value in the given equation. The left hand side (LHS) of the equation becomes: \(4 \Delta(\cot A+\cot B+\cot C)=4 \frac{1}{2}ab\sin C(\cot A+\cot B+\cot C)\).

Step 2: Apply the Trigonometric Identity

We know that \(\cot A = \frac{\cos A}{\sin A}\) and \(\sin C = \frac{c}{2R}\), where \(R\) is the circumradius of the triangle. Substituting these in the expression we obtained above, we get \(4 \frac{1}{2}ab\left(\frac{c}{2R}\right)(\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}+\frac{\cos C}{\sin C})\).

Step 3: Simplify the Expression

This simplifies to \(abc(\cot A+\cot B+\cot C)\). Also, we know \(\cot A = \frac{b^2+c^2-a^2}{2bc}\), \(\cot B = \frac{c^2+a^2-b^2}{2ac}\), and \(\cot C = \frac{a^2+b^2-c^2}{2ab}\). Replacing these into the equation we get: \(abc\left(\frac{b^2+c^2-a^2}{2bc}+\frac{c^2+a^2-b^2}{2ac}+\frac{a^2+b^2-c^2}{2ab}\right)\). The \(abc\) will cancel out and simplifying it further, the expression equals \(a^2+b^2+c^2\). So, the given LHS equals the RHS.

Key Concepts

Triangle area

The area of a triangle is a fundamental concept in geometry. It can be understood as the amount of space enclosed by the three sides of a triangle. One of the most commonly used formulas to calculate the area of a triangle is the half-product of the base and the height: \( \Delta = \frac{1}{2} \times \text{base} \times \text{height}\). However, when the height is not directly known, especially in cases of oblique triangles (triangles without a right angle), we can utilize other methods.

For any triangle with sides \( a \), \( b \), and \( c \) and one of its angles known, the trigonometric formula can be used: \( \Delta = \frac{1}{2} ab \sin C \). Here, \( a \) and \( b \) are the lengths of the two sides, and \( C \) is the angle between them. This formula is particularly helpful as it connects trigonometry and geometry together.

Understanding how to calculate the area of a triangle is crucial in various fields such as architecture, engineering, and computer graphics, where precise measurements are necessary.

Trigonometric identities

Trigonometric identities are equations involving trigonometric functions that are true for every value of the variables involved. These identities are crucial in simplifying complex trigonometric expressions and solving trigonometric equations.

Some fundamental trigonometric identities include:

• Pythagorean Identity: \( \sin^2 \theta + \cos^2 \theta = 1 \)
• Angle Sum Identities: \( \sin(A + B) = \sin A \cos B + \cos A \sin B \) and \( \cos(A + B) = \cos A \cos B - \sin A \sin B \)
• Double Angle Identities: \( \sin 2A = 2 \sin A \cos A \), \( \cos 2A = \cos^2 A - \sin^2 A \)

These identities are not only used in pure mathematics but also in various applications such as signal processing and physics. In trigonometry, identities help replace one complex expression with a simpler equivalent form, making it easier to solve equations or prove theorems. For example, the problem presented involves using the cotangent identity: \( \cot A = \frac{\cos A}{\sin A} \) to derive a proof about the properties of a triangle.

Cotangent in triangles

The cotangent of an angle in a triangle is one of the less commonly known trigonometric functions. It is the reciprocal of the tangent function, which can be represented as \( \cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta} \). While sine, cosine, and tangent are frequently used in geometry, the cotangent can often provide a helpful perspective, especially when dealing with proportions of the sides of a triangle.

In the context of triangles, cotangent is specifically useful in certain trigonometric proofs and identities. For instance, in the given exercise, the cotangents of the angles \( A \), \( B \), and \( C \) are related to the lengths of the sides of the triangle:

• \( \cot A = \frac{b^2 + c^2 - a^2}{2bc} \)
• \( \cot B = \frac{c^2 + a^2 - b^2}{2ac} \)
• \( \cot C = \frac{a^2 + b^2 - c^2}{2ab} \)

These relationships show a deeper connection between the angles and sides of triangles that can simplify the process of solving certain geometry problems or proving new theorems. Cotangent provides a unique perspective in understanding the geometric structure and symmetry of triangles.