Understanding the Range of K for tan 3A / tan A

The question asks us to find the range where the value of \(K = \frac{\tan 3A}{\tan A}\) cannot lie, given that \(\tan A\) is a real number (and not equal to 0) and \(K \neq \frac{1}{3}\).

Relating tan 3A and tan A using Trigonometry Formula

We start with the triple angle formula for tangent, which relates \(\tan 3A\) to \(\tan A\):

\(\tan 3A = \frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A}\)

We are given the equation \(K = \frac{\tan 3A}{\tan A}\). Since \(\tan A \neq 0\), we can substitute the formula for \(\tan 3A\):

\(K = \frac{\frac{3 \tan A - \tan^3 A}{1 - 3 \tan^2 A}}{\tan A}\)

\(K = \frac{\tan A (3 - \tan^2 A)}{\tan A (1 - 3 \tan^2 A)}\)

Since \(\tan A \neq 0\), we can cancel the \(\tan A\) term from the numerator and the denominator:

\(K = \frac{3 - \tan^2 A}{1 - 3 \tan^2 A}\)

Finding the Condition for Real Values of tan A

Let \(t = \tan A\). Since \(\tan A\) must be a real number, \(t\) must be real. The equation becomes:

\(K = \frac{3 - t^2}{1 - 3 t^2}\)

We need to find the values of K for which there exists a real value \(t\) satisfying this equation. Let's rearrange the equation to express \(t^2\) in terms of \(K\).

\(K(1 - 3t^2) = 3 - t^2\)

\(K - 3Kt^2 = 3 - t^2\)

\(t^2 - 3Kt^2 = 3 - K\)

\(t^2(1 - 3K) = 3 - K\)

Given that \(K \neq \frac{1}{3}\), the term \((1 - 3K)\) is not zero, so we can divide by it:

\(t^2 = \frac{3 - K}{1 - 3K}\)

For \(t = \tan A\) to be a real number, \(t^2\) must be non-negative (since the square of any real number is non-negative). Thus, we must have:

\(\frac{3 - K}{1 - 3K} \geq 0\)

Solving the Inequality for K

To solve the inequality \(\frac{3 - K}{1 - 3K} \geq 0\), we need to consider the signs of the numerator and the denominator.

• Case 1: Numerator \(\geq 0\) and Denominator > 0.
  • \(3 - K \geq 0 \implies K \leq 3\)
  • \(1 - 3K > 0 \implies 1 > 3K \implies K < \frac{1}{3}\)
  • The intersection of \(K \leq 3\) and \(K < \frac{1}{3}\) is \(K < \frac{1}{3}\).
• Case 2: Numerator \(\leq 0\) and Denominator < 0.
  • \(3 - K \leq 0 \implies K \geq 3\)
  • \(1 - 3K < 0 \implies 1 < 3K \implies K > \frac{1}{3}\)
  • The intersection of \(K \geq 3\) and \(K > \frac{1}{3}\) is \(K \geq 3\).

Combining both cases, the condition \(\frac{3 - K}{1 - 3K} \geq 0\) is satisfied when \(K \lt \frac{1}{3}\) or \(K \geq 3\).

This means that for real values of \(\tan A\), K can lie in the interval \((-\infty, \frac{1}{3}) \cup [3, \infty)\).

Identifying the Range K Cannot Lie In

The question asks for the range where K cannot lie. This is the complement of the range where K can lie.

The set of all real numbers is \((-\infty, \infty)\).

The range where K can lie is \((-\infty, \frac{1}{3}) \cup [3, \infty)\).

The complement is \((-\infty, \infty) \setminus ((-\infty, \frac{1}{3}) \cup [3, \infty))\). This complement is the open interval between \(\frac{1}{3}\) and 3, i.e., \((\frac{1}{3}, 3)\).

Therefore, for real values of \(\tan A\), K cannot lie strictly between \(\frac{1}{3}\) and 3.

Comparing with Options

Let's look at the given options, which are intervals:

• Option 1: between \(\frac{1}{2}\) and 2, i.e., \((\frac{1}{2}, 2)\).
• Option 2: between \(\frac{1}{3}\) and 3, i.e., \((\frac{1}{3}, 3)\).
• Option 3: between \(\frac{1}{5}\) and 5, i.e., \((\frac{1}{5}, 5)\).
• Option 4: between \(\frac{1}{7}\) and 7, i.e., \((\frac{1}{7}, 7)\).

The range where K cannot lie is \((\frac{1}{3}, 3)\), which exactly matches Option 2.

The final answer is the range between \(\frac{1}{3}\) and 3.

Revision Table: Key Concepts

Additional Information: Trigonometric Identities and Ranges

Trigonometric identities are crucial for simplifying expressions and solving equations in trigonometry. The triple angle formulas are derived from the sum and double angle formulas.

• Sum Formula: \(\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\)
• Double Angle Formula: \(\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}\) (derived by setting B=A in the sum formula)
• Triple Angle Formula: \(\tan 3A = \tan(2A+A) = \frac{\tan 2A + \tan A}{1 - \tan 2A \tan A}\). Substituting the formula for \(\tan 2A\) leads to the expression used in the problem.

The range of the tangent function for real angles (where it is defined) is \((-\infty, \infty)\). This means that for any real number \(t\), there exists a real angle A such that \(\tan A = t\), provided A is not of the form \((n + \frac{1}{2})\pi\) for integer n. The condition that \(\tan A\) is a real number is always true when A is a real angle, except at the asymptotes of the tan function.

In this problem, the condition for \(\tan A\) to be real translates to \((\tan A)^2 \geq 0\), which means the expression for \(t^2 = (\tan A)^2\) in terms of K must be non-negative. This is a standard technique to find the range of a parameter when one variable is expressed as a function of another, and constraints on the variables are given.