Trigonometry Problem: Solving for p² - q²

This problem involves manipulating given trigonometric equations to find the value of an expression. We are given two equations relating variables \(p\) and \(q\) with trigonometric functions \(\cot \theta\) and \(\operatorname{cosec} \theta\), and constants \(a\) and \(b\).

Understanding the Given Trigonometric Equations

The two equations provided are:

1. \(a \cot \theta + b \operatorname{cosec} \theta = p\)
2. \(b \cot \theta + a \operatorname{cosec} \theta = q\)

We need to find the value of \(p^2 - q^2\).

Calculating p² and q²

First, let's square both equations to find \(p^2\) and \(q^2\).

Calculating p²:

Squaring the first equation \((a \cot \theta + b \operatorname{cosec} \theta = p)\):

\(p^2 = (a \cot \theta + b \operatorname{cosec} \theta)^2\)

Using the \((x+y)^2 = x^2 + y^2 + 2xy\) formula:

\(p^2 = (a \cot \theta)^2 + (b \operatorname{cosec} \theta)^2 + 2(a \cot \theta)(b \operatorname{cosec} \theta)\)

\(p^2 = a^2 \cot^2 \theta + b^2 \operatorname{cosec}^2 \theta + 2ab \cot \theta \operatorname{cosec} \theta \quad \text{(Equation 3)}\)

Calculating q²:

Squaring the second equation \((b \cot \theta + a \operatorname{cosec} \theta = q)\):

\(q^2 = (b \cot \theta + a \operatorname{cosec} \theta)^2\)

Using the \((x+y)^2 = x^2 + y^2 + 2xy\) formula:

\(q^2 = (b \cot \theta)^2 + (a \operatorname{cosec} \theta)^2 + 2(b \cot \theta)(a \operatorname{cosec} \theta)\)

\(q^2 = b^2 \cot^2 \theta + a^2 \operatorname{cosec}^2 \theta + 2ab \cot \theta \operatorname{cosec} \theta \quad \text{(Equation 4)}\)

Finding p² - q²

Now, we subtract Equation 4 from Equation 3:

\(p^2 - q^2 = (a^2 \cot^2 \theta + b^2 \operatorname{cosec}^2 \theta + 2ab \cot \theta \operatorname{cosec} \theta) - (b^2 \cot^2 \theta + a^2 \operatorname{cosec}^2 \theta + 2ab \cot \theta \operatorname{cosec} \theta)\)

Distribute the negative sign:

\(p^2 - q^2 = a^2 \cot^2 \theta + b^2 \operatorname{cosec}^2 \theta + 2ab \cot \theta \operatorname{cosec} \theta - b^2 \cot^2 \theta - a^2 \operatorname{cosec}^2 \theta - 2ab \cot \theta \operatorname{cosec} \theta\)

Notice that the terms \(+ 2ab \cot \theta \operatorname{cosec} \theta\) and \(- 2ab \cot \theta \operatorname{cosec} \theta\) cancel each other out.

\(p^2 - q^2 = a^2 \cot^2 \theta + b^2 \operatorname{cosec}^2 \theta - b^2 \cot^2 \theta - a^2 \operatorname{cosec}^2 \theta\)

Simplifying the Expression using Trigonometric Identity

Rearrange the terms to group those with \(a^2\) and \(b^2\):

\(p^2 - q^2 = (b^2 \operatorname{cosec}^2 \theta - a^2 \operatorname{cosec}^2 \theta) + (a^2 \cot^2 \theta - b^2 \cot^2 \theta)\)

Factor out common terms:

\(p^2 - q^2 = \operatorname{cosec}^2 \theta (b^2 - a^2) + \cot^2 \theta (a^2 - b^2)\)

To use the identity \(\operatorname{cosec}^2 \theta - \cot^2 \theta = 1\), we need the terms inside the parentheses to be the same. We can rewrite \((a^2 - b^2)\) as \(-(b^2 - a^2)\):

\(p^2 - q^2 = \operatorname{cosec}^2 \theta (b^2 - a^2) + \cot^2 \theta (-(b^2 - a^2))\)

\(p^2 - q^2 = \operatorname{cosec}^2 \theta (b^2 - a^2) - \cot^2 \theta (b^2 - a^2)\)

Now, factor out the common term \((b^2 - a^2)\):

\(p^2 - q^2 = (b^2 - a^2) (\operatorname{cosec}^2 \theta - \cot^2 \theta)\)

Recall the fundamental trigonometric identity: \(\operatorname{cosec}^2 \theta - \cot^2 \theta = 1\).

Substitute this into the expression:

\(p^2 - q^2 = (b^2 - a^2)(1)\)

\(p^2 - q^2 = b^2 - a^2\)

Final Answer

The value of \(p^2 - q^2\) is \(b^2 - a^2\).

Revision Table: Key Trigonometric Identities and Concepts

Additional Information: Understanding Reciprocal Identities

The functions \(\cot \theta\) and \(\operatorname{cosec} \theta\) are related to the basic trigonometric functions \(\sin \theta\) and \(\cos \theta\) through reciprocal identities:

• \(\cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}\)
• \(\operatorname{cosec} \theta = \frac{1}{\sin \theta}\)

The identity \(\operatorname{cosec}^2 \theta - \cot^2 \theta = 1\) can be derived from the identity \(\sin^2 \theta + \cos^2 \theta = 1\) by dividing all terms by \(\sin^2 \theta\) (assuming \(\sin \theta \ne 0\)):

\(\frac{\sin^2 \theta}{\sin^2 \theta} + \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{1}{\sin^2 \theta}\)

\(1 + \cot^2 \theta = \operatorname{cosec}^2 \theta\)

Rearranging this gives: \(\operatorname{cosec}^2 \theta - \cot^2 \theta = 1\). This identity is fundamental in solving trigonometric problems involving \(\cot\) and \(\operatorname{cosec}\).