Problem 62 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

In a triangle \(P Q R, \angle R=\frac{\pi}{2} .\) If \(\tan \left(\frac{P}{2}\right)\) and \(\tan \left(\frac{Q}{2}\right)\) are the roots of \(a x^{2}+b x+c=0, a \neq 0\) then \(\quad\) (A) \(a=b+c\) (B) \(c=a+b\) (C) \(b=c\) (D) \(b=a+c\)

Step-by-Step Solution

Answer

(C) \(b = c\)

Step 1: Use Triangle Properties

Since \(\angle R = \frac{\pi}{2}\), triangle \(PQR\) is a right triangle. The angles \(P\) and \(Q\) must satisfy \(P + Q = \frac{\pi}{2}\).

Step 2: Apply Half-Angle Identity

The roots of the quadratic \(ax^2 + bx + c = 0\) are \(\tan \left(\frac{P}{2}\right)\) and \(\tan \left(\frac{Q}{2}\right)\). For a right triangle, \(\tan\left(\frac{P}{2}\right)\) and \(\tan\left(\frac{Q}{2}\right)\) relate to the half-angle identities \(\tan\left(\frac{P}{2}\right)\tan\left(\frac{Q}{2}\right) = 1\).

Step 3: Use Relationship of Quadratic Roots

For roots \(p\) and \(q\) of a quadratic equation \(ax^2 + bx + c = 0\), the relationships are \(p + q = -\frac{b}{a}\) and \(pq = \frac{c}{a}\). Given that \(pq = 1\) from the half-angle identity, \(\frac{c}{a} = 1\), making \(c = a\).

Step 4: Handle Sum of Roots Equation

Given \(p + q = -\frac{b}{a}\) and knowing \(p\) and \(q\) satisfy the right triangle relationship \(p + q = -1\) (since \(p + q\) must be \(-1\) for the roots of the form \(\tan\left(\frac{P}{2}\right), \tan\left(\frac{Q}{2}\right)\)), we equate this to \(-\frac{b}{a}\) leading to \(b = a\).

Step 5: Combine Relationships

Given \(c = a\) and \(b = a\), substitute these relationships into the options: realize that in option \(b = c\), this relation holds true since both \(b\) and \(c\) equal \(a\).

Key Concepts

Quadratic Equations

Quadratic equations are mathematical expressions of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). Solving them involves finding the values of \(x\) that make the equation true. These values are known as the roots. A fundamental property of quadratic equations is that they can have up to two roots, which can be real or complex. There are several methods to solve quadratic equations, including:

• The quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• Factoring, when the equation can be neatly expressed as a product of binomials.
• Completing the square, transforming the equation into a perfect square form.

For a specific quadratic equation, like in our triangle problem, the roots \(p\) and \(q\) are related directly to the coefficients. The sum of the roots \(p + q\) equals \(-\frac{b}{a}\), and the product of the roots \(pq\) equals \(\frac{c}{a}\). Understanding these relationships is crucial when dealing with the properties of quadratic roots.

Half-Angle Identities

Half-angle identities are trigonometric relationships that relate the tangent of half an angle to the sides of the triangle. They are particularly useful in problems involving right triangles. In our exercise, we are given that \(\tan\left(\frac{P}{2}\right)\) and \(\tan\left(\frac{Q}{2}\right)\) are the roots of the quadratic equation.

• The half-angle identity for tangent is: \(\tan\left(\frac{\theta}{2}\right) = \pm\sqrt{\frac{1 - \cos\theta}{1 + \cos\theta}}\).
• For a right triangle where one angle is \(\frac{\pi}{2}\), the other angles \(P\) and \(Q\) sum up to the remaining \(\frac{\pi}{2}\), making half-angle identities very practical.

In the context of the quadratic equation, these identities provide insight into the relationship between the angles. Notably, the product of the two tangent half-angles equates to 1 if the quadrants align favorably in a right triangle. This ensures that \(\tan\left(\frac{P}{2}\right)\tan\left(\frac{Q}{2}\right) = 1\), linking directly to our problem's given condition \(pq = 1\).

Triangle Properties

Triangles, especially right triangles, have specific properties that are essential in understanding problems involving trigonometry. A right triangle is one where one angle is exactly \(90^\circ\), or \(\frac{\pi}{2}\) radians. This makes the sum of the other two angles equal to \(\frac{\pi}{2}\), fulfilling the angle sum condition of all triangles, which is \(\pi\) radians.

• In the context of right triangles, the Pythagorean theorem holds: \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse.
• Trigonometric ratios like sine, cosine, and tangent are based on the triangle's angles and side lengths, providing powerful tools for solving geometric problems.

For our right triangle \(PQR\), knowing that \(\angle R = \frac{\pi}{2}\) helps establish the foundation to apply trigonometric identities effectively. The fact that \(P + Q = \frac{\pi}{2}\) allows us to use the tangent half-angle identities to reflect the roots of the quadratic. These triangle properties give us the structure necessary to solve the problem and understand the relationship between the angles and the quadratic equation roots.