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In a triangle \(\mathrm{PQR}, \angle R=\frac{\pi}{2}\). 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: [AIEEE - 2005] (a) \(b=a+c\) (b) \(b=c\) (c) \(c=a+b\) (d) \(a=b+c\)

Short Answer

Expert verified
The answer is (d) \( a = b + c \).

Step by step solution

01

Identify Key Information Given in the Problem

We have a right triangle \( \triangle PQR \) where \( \angle R = \frac{\pi}{2} \). We need to determine a relationship between the coefficients \(a, b,\) and \(c\) of the quadratic equation whose roots are \(\tan\left(\frac{P}{2}\right)\) and \(\tan\left(\frac{Q}{2}\right)\).
02

Use the Half-Angle Identity for Tangent

For a right triangle, if \( \angle R = \frac{\pi}{2} \), \( P + Q = \pi/2 \). The half-angle tangent identities are: \[ \tan\left(\frac{P}{2}\right) = \sqrt{\frac{1 - \cos P}{1 + \cos P}} \] and \[ \tan\left(\frac{Q}{2}\right) = \sqrt{\frac{1 - \cos Q}{1 + \cos Q}} \].
03

Relate Angles P and Q in Terms of Triangle Sides

Using the triangle sides, \( \tan P = \frac{QR}{PR} \) and \( \tan Q = \frac{PR}{QR} \). Also, using \( P + Q = \frac{\pi}{2} \), then \( \tan Q = \cot P \). Hence, we have \( \tan Q \cdot \tan P = 1 \).
04

Quadratic Equation Using Product of Roots Identity

For roots \( \tan\left(\frac{P}{2}\right) \) and \( \tan\left(\frac{Q}{2}\right) \), the product is given by the quadratic identity: \[ \tan\left(\frac{P}{2}\right) \times \tan\left(\frac{Q}{2}\right) = \frac{1 - \frac{\cos P}{\cos Q}}{1 + \frac{\cos P}{\cos Q}} = \frac{PQ}{PR} \times \frac{PR}{PQ} = 0 \].
05

Apply Viète's Formulas

According to Viète's formulas for the quadratic equation \(ax^2 + bx + c = 0\), the sum of the roots \(s = \tan\left(\frac{P}{2}\right) + \tan\left(\frac{Q}{2}\right) = -\frac{b}{a}\) and the product of the roots \(p = \tan\left(\frac{P}{2}\right) \cdot \tan\left(\frac{Q}{2}\right) = \frac{c}{a}\). Given the derived product in Step 4, \( p = 0 \), so \( c = 0 \). Unfortunately, since both roots are zero, this term is eliminated.
06

Interpret and Choose the Correct Option

From the product \(\tan\left(\frac{P}{2}\right) \cdot \tan\left(\frac{Q}{2}\right) = 0\), which implies \( c = 0 \), the potential relation is \(b = a\), leaving \(b = a + c\) where \(c = 0\), hence \(b = a + 0 = a\). Thus the correct choice is \( b = a + c \). Hence, the selected option is \(d.\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Right Triangle Properties
A right triangle is a special kind of triangle where one angle is exactly 90 degrees or \( \frac{\pi}{2} \) radians. In such triangles, a few properties help us understand their geometry better:
  • **Pythagorean Theorem**: This states that for a right triangle with sides \( a \), \( b \), and hypotenuse \( c \), the relationship is \( a^2 + b^2 = c^2 \).
  • **Complementary Angles**: The two non-right angles, let's call them \( P \) and \( Q \), are complementary, meaning \( P + Q = \frac{\pi}{2} \) or \( 90^\circ \).
  • **Trigonometric Ratios**: These triangles are well-known for their use in defining sine, cosine, and tangent. The tangent of an angle, like \( P \), can be defined as \( \tan P = \frac{\text{opposite}}{\text{adjacent}} \). Therefore, \( \tan P \) and \( \tan Q \) are inverses since \( \tan Q = \cot P \).
A keen understanding of these properties allows us to manipulate right triangles for solving problems related to angles and side lengths. This concept is crucial when dealing with half-angle identities as seen in trigonometry problems.
Quadratic Equations
Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \), where \( a eq 0 \). These are solved to find the values of \( x \) that satisfy the equation. Here are a few key points about quadratic equations:
  • **Standard Form**: The typical form is \( ax^2 + bx + c = 0 \), and it's polynomial with a degree of 2.
  • **Roots**: The values of \( x \) which satisfy the equation are called the roots, which might be real or complex.
  • **Factoring Method**: Sometimes, quadratics can be factorized into two binomials which, when set to zero, can yield the roots.
  • **Quadratic Formula**: For quadratics that are not easily factored, the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) can be used to find the roots.
Quadratic equations are particularly significant when dealing with problems requiring symmetrical functions and equations, such as those involving roots like \( \tan(\frac{P}{2}) \) and \( \tan(\frac{Q}{2}) \). Understanding quadratics is crucial for applying Viète's formulas, which delve into the sum and product of roots.
Viète's Formulas
Viète's formulas provide a deep insight into the relationship between the coefficients and the roots of a polynomial equation. For a quadratic equation \( ax^2 + bx + c = 0 \):
  • **Sum of Roots**: If \( \alpha \) and \( \beta \) are the roots, then their sum is given by \( \alpha + \beta = -\frac{b}{a} \).
  • **Product of Roots**: The product of the roots is \( \alpha \beta = \frac{c}{a} \).
These identities are directly applied to understand problems where determining the sum or product of certain calculated routines (like half-angle tangents) is necessary. For example, if we derive the identity that \( \tan(\frac{P}{2}) \cdot \tan(\frac{Q}{2}) = 0 \), using Viète's product formula, we infer that \( c = 0 \), aligning with certain coefficient conditions like \( b = a + c \). Understanding Viète's formulas allows for a strategic unpacking of even more complex polynomial conditions beneath surface-level calculations.

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