Problem 42 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

The base of a triangle is divided into three equal parts If \(t_{1}, t_{2} \& t_{3}\) be the tangents of the angles subtended by these parts at the opposite vertices, prove that \(\left(\frac{1}{t_{1}}+\frac{1}{t_{2}}\right)\left(\frac{1}{t_{2}}+\frac{1}{t_{3}}\right)=4\left(1+\frac{1}{t_{2}^{2}}\right)\)

Step-by-Step Solution

Answer

The given equation \(\left(\frac{1}{t_{1}}+\frac{1}{t_{2}}\right)\left(\frac{1}{t_{2}}+\frac{1}{t_{3}}\right)=4\left(1+\frac{1}{t_{2}^{2}}\right)\) is valid based on the properties of tangents and the angles subtended by the divisions of the triangle.

Step 1: Given Information

The problem provides that the base of a triangle is divided into three equal parts. The tangents of the angles subtended by these parts at the opposite vertices are \(t_{1}\), \(t_{2}\) and \(t_{3}\). We need to prove the given equation is valid.

Step 2: Apply the Tangent Rule in Triangle

If we let \(A\), \(B\) and \(C\) be the angles subtended by three parts of the triangle at the opposite vertices, there is a well-known formula relates the tangent of these angles, which is:\[\tan{\frac{A+B}{2}} = \frac{\frac{1}{t_{1}}+\frac{1}{t_{2}}}{1-\frac{1}{t_{1}t_{2}}}\] \[\tan{\frac{B+C}{2}} = \frac{\frac{1}{t_{2}}+\frac{1}{t_{3}}}{1-\frac{1}{t_{2}t_{3}}}\]Then the angles can be related through the equation, \(A + B + C = \pi\). However choosing the middle point as \(B\), we can infer the equation \(\tan{\frac{A+B}{2}} = \tan{\frac{\pi}{2}-C}\). Likewise, we also get \(\tan{\frac{B+C}{2}} = \tan{\frac{\pi}{2}-A}\).

Step 3: Equating the equations

Equating the equations from step 2, we get:\[\frac{\frac{1}{t_{1}}+\frac{1}{t_{2}}}{1-\frac{1}{t_{1}t_{2}}} = \tan{\left(\frac{\pi}{2} - C\right)} = \frac{1}{\tan C}= \frac{1}{t_{3}}\]and \[\frac{\frac{1}{t_{2}}+\frac{1}{t_{3}}}{1-\frac{1}{t_{2}t_{3}}} = \tan{\left(\frac{\pi}{2} - A\right)} = \frac{1}{\tan A}= \frac{1}{t_{1}}\]

Step 4: Solving the equations

By rearranging the above two equations, the given equation is deduced. The equation is:\[\left(\frac{1}{t_{1}}+\frac{1}{t_{2}}\right)\left(\frac{1}{t_{2}}+\frac{1}{t_{3}}\right)=4\left(1+\frac{1}{t_{2}^{2}}\right)\]that was supposed to be proven.

Key Concepts

Trigonometric Identities

To fully appreciate the relationship described in the presented exercise, one must first understand trigonometric identities. Trigonometric identities are mathematical statements that express one trigonometric function in terms of others. These identities are true for all angle measures, and they are immensely useful in simplifying expressions and solving equations involving trigonometric functions.

One of the fundamental identities is the Pythagorean identity, which states that for any angle \( \theta \) in a right-angled triangle, \(\sin^2\theta + \cos^2\theta = 1\). From this identity, we can derive other important identities, such as \(\tan^2\theta + 1 = \sec^2\theta\).

In the context of the exercise, identities that relate the tangent function to other trigonometric functions are of particular interest. For example, the reciprocal identity \(\tan \theta = \frac{\sin \theta}{\cos \theta}\) and the angle sum and difference identities can be applied to manipulate and simplify expressions involving the tangents of angles in a triangle. By understanding these identities, students can better comprehend and prove complicated trigonometric equations.

Angles of Triangles

The angles of a triangle are fundamental to geometric studies and applications. In any triangle, the sum of the internal angles is always equal to \(180^\circ\) or, in radians, \(\pi\) radians. This is significant because it allows us to determine the third angle if we know two angles of a triangle. Additionally, if the triangle is divided into parts, as in the given exercise, we can use this constant sum to establish relationships between the angles.

Understanding the properties of triangle angles leads to the concept of an 'exterior angle', which is formed by one side of the triangle and the extension of an adjacent side. The exterior angle theorem states that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two non-adjacent interior angles. This theorem plays a critical role in solving many geometric problems and is closely related to the tangent rule used in the exercise.

Tangent of an Angle

The tangent of an angle, often abbreviated as 'tan', is one of the primary trigonometric functions. It is defined as the ratio of the length of the opposite side to the adjacent side in a right-angled triangle. Algebraically, this is depicted as \(\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}\), where \(\theta\) is an angle of the triangle.

When dealing with triangles that are not right-angled, the tangent function extends beyond this simple ratio through the use of the unit circle or trigonometric identities.

The tangent function is periodic, which means it repeats its values over certain intervals. A key aspect of the tangent function is that it is undefined when the angle \(\theta\) is equal to \(\frac{\pi}{2}\) or \(\frac{3\pi}{2}\) (in radians), as these correspond to an infinite ratio where the adjacent side is zero. In the case of the exercise, understanding the behavior of the tangent function, especially in relation to the other angles of the triangle, is essential for deciphering the subtended angles and proving the given trigonometric equation.