Analyzing the Trigonometric Problem

We are given two trigonometric expressions involving \(\theta\): The goal is to find the value of the expression \(p^2q^2 (p^2 + q^2+3)\). We need to simplify \(p\) and \(q\) first using fundamental trigonometric identities.

1. \(p = \text{cosec}\theta - \sin\theta\)
2. \(q = \sec\theta - \cos\theta\)

Step-by-Step Simplification

Step 1: Express p in terms of sin \(\theta\) and cos \(\theta\).

Recall that \(\text{cosec}\theta = \frac{1}{\sin\theta}\). Substituting this:

\(p = \frac{1}{\sin\theta} - \sin\theta\)

To combine these terms, we find a common denominator:

\(p = \frac{1 - \sin^2\theta}{\sin\theta}\)

Using the Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\), we know that \(1 - \sin^2\theta = \cos^2\theta\). So,

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

Step 2: Express q in terms of sin \(\theta\) and cos \(\theta\).

Recall that \(\sec\theta = \frac{1}{\cos\theta}\). Substituting this:

\(q = \frac{1}{\cos\theta} - \cos\theta\)

Finding a common denominator:

\(q = \frac{1 - \cos^2\theta}{\cos\theta}\)

Using the Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\), we know that \(1 - \cos^2\theta = \sin^2\theta\). So,

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

Step 3: Calculate \(p^2\) and \(q^2\).

Squaring the expressions for \(p\) and \(q\):

\(p^2 = \left(\frac{\cos^2\theta}{\sin\theta}\right)^2 = \frac{\cos^4\theta}{\sin^2\theta}\)

\(q^2 = \left(\frac{\sin^2\theta}{\cos\theta}\right)^2 = \frac{\sin^4\theta}{\cos^2\theta}\)

Step 4: Calculate the product \(p^2q^2\).

Multiply \(p^2\) and \(q^2\):

\(p^2q^2 = \left(\frac{\cos^4\theta}{\sin^2\theta}\right) \times \left(\frac{\sin^4\theta}{\cos^2\theta}\right)\)

Simplify by cancelling terms:

\(p^2q^2 = \frac{\cos^4\theta \cdot \sin^4\theta}{\sin^2\theta \cdot \cos^2\theta} = \cos^2\theta \sin^2\theta\)

Step 5: Calculate \(p^2 + q^2\).

Add \(p^2\) and \(q^2\):

\(p^2 + q^2 = \frac{\cos^4\theta}{\sin^2\theta} + \frac{\sin^4\theta}{\cos^2\theta}\)

Find a common denominator (\(\sin^2\theta \cos^2\theta\)):

\(p^2 + q^2 = \frac{\cos^4\theta \cdot \cos^2\theta + \sin^4\theta \cdot \sin^2\theta}{\sin^2\theta \cos^2\theta} = \frac{\cos^6\theta + \sin^6\theta}{\sin^2\theta \cos^2\theta}\)

We use the identity \(a^3 + b^3 = (a+b)(a^2-ab+b^2)\). Let \(a = \cos^2\theta\) and \(b = \sin^2\theta\). Then \(a+b = \cos^2\theta + \sin^2\theta = 1\).

\(\cos^6\theta + \sin^6\theta = (\cos^2\theta)^3 + (\sin^2\theta)^3\)

\(= (\cos^2\theta + \sin^2\theta)((\cos^2\theta)^2 - \cos^2\theta \sin^2\theta + (\sin^2\theta)^2)\)

\(= (1)(\cos^4\theta - \cos^2\theta \sin^2\theta + \sin^4\theta)\)

Also, \(\cos^4\theta + \sin^4\theta = (\cos^2\theta + \sin^2\theta)^2 - 2\cos^2\theta \sin^2\theta = 1^2 - 2\cos^2\theta \sin^2\theta = 1 - 2\cos^2\theta \sin^2\theta\).

Substituting this back:

\(\cos^6\theta + \sin^6\theta = (1 - 2\cos^2\theta \sin^2\theta) - \cos^2\theta \sin^2\theta = 1 - 3\cos^2\theta \sin^2\theta\)

Now substitute this result back into the expression for \(p^2 + q^2\):

\(p^2 + q^2 = \frac{1 - 3\cos^2\theta \sin^2\theta}{\sin^2\theta \cos^2\theta}\)

Step 6: Evaluate the target expression \(p^2q^2 (p^2 + q^2+3)\).

First, let's find \(p^2 + q^2 + 3\):

\(p^2 + q^2 + 3 = \left(\frac{1 - 3\cos^2\theta \sin^2\theta}{\sin^2\theta \cos^2\theta}\right) + 3\)

Combine the terms using the common denominator \(\sin^2\theta \cos^2\theta\):

\(p^2 + q^2 + 3 = \frac{1 - 3\cos^2\theta \sin^2\theta + 3(\sin^2\theta \cos^2\theta)}{\sin^2\theta \cos^2\theta}\)

\(p^2 + q^2 + 3 = \frac{1 - 3\cos^2\theta \sin^2\theta + 3\cos^2\theta \sin^2\theta}{\sin^2\theta \cos^2\theta}\)

\(p^2 + q^2 + 3 = \frac{1}{\sin^2\theta \cos^2\theta}\)

Now, substitute the values of \(p^2q^2\) and \((p^2 + q^2 + 3)\) into the expression we need to evaluate:

\(p^2q^2 (p^2 + q^2+3) = (\cos^2\theta \sin^2\theta) \times \left(\frac{1}{\sin^2\theta \cos^2\theta}\right)\)

\(p^2q^2 (p^2 + q^2+3) = 1\)

Final Answer Verification

The calculations show that the expression \(p^2q^2 (p^2 + q^2+3)\) simplifies to 1, regardless of the value of \(\theta\) (as long as \(\sin\theta \neq 0\) and \(\cos\theta \neq 0\)). This matches one of the given options.