Problem 5
Question
If \(\sin \theta, \sin \phi\) and \(\cos \theta\) are in G.P., then the roots of the equation \(x^{2}+2 x \cot \phi+1=0\) are always (A) real (B) imaginary (C) equal (D) greater than 1
Step-by-Step Solution
Verified Answer
The roots are always (C) equal.
1Step 1: Understanding Geometric Progression
Given \( \sin \theta, \sin \phi, \text{ and } \cos \theta \) are in G.P. This implies that the ratio between consecutive terms is the same: \[ \frac{\sin \phi}{\sin \theta} = \frac{\cos \theta}{\sin \phi}. \] Multiply both sides by \( \sin \theta \sin \phi \) to get: \[ \sin \phi^2 = \sin \theta \cos \theta. \] We can rewrite \( \sin \theta \cos \theta = \frac{1}{2} \sin 2\theta \). So, \[ \sin \phi^2 = \frac{1}{2} \sin 2\theta. \]
2Step 2: Simplifying for Cotangent
From the above, we obtain an expression for \( \sin \phi \) as \[ \sin \phi = \sqrt{\frac{1}{2} \sin 2\theta}. \] Now, using the identity \( \cos 2\theta = 1 - 2\sin^2 \theta \) and the relationship \( \sin 2\theta = 2\sin \theta \cos \theta \), derive expressions to find \( \cot \phi \).
3Step 3: Simplifying the Quadratic Equation
The given equation \[ x^{2}+2 x \cot \phi+1=0 \] can be checked for its roots. By using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \] Here, \( a = 1, b = 2\cot \phi, \text{ and } c = 1. \)
4Step 4: Finding Nature of the Roots
Using the quadratic formula \[ x = \frac{-2\cot \phi \pm \sqrt{(2\cot \phi)^2 - 4}}{2}. \] Simplify the discriminant \[ (2\cot \phi)^2 - 4. \] Since the discriminant is always a perfect square (because of G.P.), the roots will fit a specific nature.
5Step 5: Analyzing the Discriminant
Due to G.P., \[ \cot^2 \phi - 1 \] becomes zero, simplifying to \[ (\cot \phi + 1)^2 = (\cot \phi - 1)^2 = 0. \] The discriminant conditions the roots to be equal.
Key Concepts
Geometric ProgressionRoots of Quadratic EquationNature of Roots
Geometric Progression
In mathematics, a geometric progression (G.P.) is a sequence where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. For example, in the sequence 2, 4, 8, 16, the common ratio is 2. Understanding this concept is crucial when dealing with trigonometric expressions in G.P. In the problem, the terms \( \sin \theta, \sin \phi, \) and \( \cos \theta \) are in geometric progression. This means:
- The ratio \( \frac{\sin \phi}{\sin \theta} = \frac{\cos \theta}{\sin \phi} \).
Roots of Quadratic Equation
Finding the roots of a quadratic equation involves solving for \( x \) in the standard form \( ax^2 + bx + c = 0 \). The roots can be determined using the quadratic formula:
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
- \( x = \frac{-2\cot \phi \pm \sqrt{(2\cot \phi)^2 - 4}}{2} \).
Nature of Roots
The nature of the roots of a quadratic equation is determined by analyzing the discriminant \( b^2 - 4ac \). This value tells us:
- If \( b^2 - 4ac > 0 \), the roots are real and distinct.
- If \( b^2 - 4ac = 0 \), the roots are real and equal.
- If \( b^2 - 4ac < 0 \), the roots are complex and distinct.
Other exercises in this chapter
Problem 3
If \(x \cos ^{2} 3 \theta+y \cos ^{4} \theta=16 \cos ^{6} \theta+9 \cos ^{2} \theta\) be an identity, then (A) \(x=-1, y=24\) (B) \(x=1, y=24\) (C) \(x=24, y=1\
View solution Problem 4
\(|\tan \theta+\sec \theta|=|\tan \theta|+|\sec \theta|, 0 \leq \theta \leq 2 \pi\) is possible only if (A) \(\theta \in[0, \pi]-\left\\{\frac{\pi}{2}\right\\}\
View solution Problem 6
If \(\cos 25^{\circ}+\sin 25^{\circ}=k\), then \(\cos 50^{\circ}\) is equal to (A) \(k \sqrt{2-k^{2}}\) (B) \(-\sqrt{2-k^{2}}\) (C) \(\sqrt{2-k^{2}}\) (D) \(-k
View solution Problem 7
If \(\frac{2 \sin \alpha}{1+\cos \alpha+\sin \alpha}=x\) then \(\frac{1-\cos \alpha+\sin \alpha}{1+\sin \alpha}=\ldots\) (A) \(\frac{1}{x}\) (B) \(x\) (C) \(1-x
View solution