Understanding the Problem: Trigonometric Expressions

The problem asks us to find the value of the product \(pq\), where \(p\) and \(q\) are given trigonometric expressions involving \(\tan x\) and \(\tan 3x\). The expressions are defined for \(0 < x < \pi\), with the exclusion \(x \neq \frac{\pi}{2}\).

The given expressions are:

• \(p = \frac{1}{3} - \frac{\tan 3x}{\tan x}\)
• \(q = 1 - 3 \tan^2 x\)

We need to simplify the expression for \(p\) first, using a trigonometric identity for \(\tan 3x\).

Simplifying the Expression for p

We use the triple angle formula for tangent, which is:

Substitute this formula into the expression for \(p\):

To simplify the fraction in the second term, we can rewrite it as:

Factor out \(\tan x\) from the numerator:

Given the domain \(0 < x < \pi\) and \(x \neq \frac{\pi}{2}\), \(\tan x\) is generally non-zero (except possibly at \(x=\pi\), but the interval is open). If \(\tan x \neq 0\), we can cancel \(\tan x\) from the numerator and the denominator:

So, the expression for \(p\) becomes:

Now, combine the terms by finding a common denominator, which is \(3 (1 - 3 \tan^2 x)\):

Combine like terms in the numerator:

Note that the denominator \(1 - 3 \tan^2 x\) cannot be zero for the original expression for \(p\) to be well-defined after cancellation of \(\tan x\) in the way we did, and for the final answer to be a constant other than 0 (as seen from the options). If \(1 - 3 \tan^2 x = 0\), then \(q = 0\), and thus \(pq = 0\), which is not among the options. Therefore, \(1 - 3 \tan^2 x \neq 0\).

Calculating the Product pq

Now we multiply the simplified expression for \(p\) by the expression for \(q\):

Since \(1 - 3 \tan^2 x \neq 0\), we can cancel the term \((1 - 3 \tan^2 x)\) from the numerator and the denominator:

Conclusion

The product \(pq\) simplifies to a constant value, \(-\frac{8}{3}\).

Revision Table: Key Concepts

Additional Information: Domain Restrictions and Trigonometric Expressions

The problem specifies the domain \(0 < x < \pi, x \neq \frac{\pi}{2}\). Let's consider why these restrictions are important for the given trigonometric expressions.

• \(\tan x\) is undefined at \(x = \frac{\pi}{2}\) within the interval \(0 < x < \pi\). This is explicitly excluded in the domain.
• \(\tan 3x\) involves the denominator \(1 - 3 \tan^2 x\). If \(1 - 3 \tan^2 x = 0\), then \(\tan^2 x = \frac{1}{3}\), which means \(\tan x = \pm \frac{1}{\sqrt{3}}\). Within the interval \(0 < x < \pi\), this occurs at \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\). At these values, \(\tan 3x\) is undefined.
• The expression for \(p\) involves \(\frac{\tan 3x}{\tan x}\). This expression is undefined if either \(\tan x = 0\) or \(\tan 3x\) is undefined. \(\tan x = 0\) at \(x=\pi\) (not in interval) and multiples of \(\pi\). \(\tan x \neq 0\) for \(0 < x < \pi, x \neq \frac{\pi}{2}\). \(\tan 3x\) is undefined at \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\) in the interval.
• The expression for \(q\) is \(1 - 3 \tan^2 x\), which is defined for all \(x\) where \(\tan x\) is defined. However, as seen in the simplification of \(p\), the term \(1 - 3 \tan^2 x\) appears in the denominator of the simplified \(p\). For the cancellation step to be valid and the final product \(pq\) to be \(-\frac{8}{3}\), we require \(1 - 3 \tan^2 x \neq 0\), which means \(x \neq \frac{\pi}{6}\) and \(x \neq \frac{5\pi}{6}\). The problem does not explicitly exclude these values, but the fact that the answer is a constant (and one of the options) suggests that the expression \(pq\) must be equal to this constant value for all valid \(x\) in the domain, implying that the intermediate cancellation is valid. If \(x = \frac{\pi}{6}\) or \(x = \frac{5\pi}{6}\), then \(q=0\), and \(p\) is undefined due to \(\tan 3x\) being undefined. In a typical exam scenario where a constant answer is expected, we proceed with the simplification assuming the expression is well-behaved for the intended scope of the question.