What is $p^2q^2 (p^2 + q^2+3)$ equal to?

Question

Prepp Team · Accepted Answer

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.

$p = \text{cosec}\theta - \sin\theta$
$q = \sec\theta - \cos\theta$

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.