If `(tan 3A)/(tan A)=k`, show that `( sin 3A)/(sin A)= (2k)/(k-1)` and hence or otherwise prove that either `k gt 3` or `k lt 1/3`.

Answer

To solve the problem, we start with the given equation:

\[ \frac{\tan 3A}{\tan A} = k \]

### Step 1: Express \(\tan 3A\) in terms of \(\tan A\)

Using the tangent triple angle formula:

\[ \tan 3A = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A} \]

Substituting this into the given equation:

\[ \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A} = k \]

### Step 2: Clear the fraction

Multiply both sides by \(1 - 3\tan^2 A\):

\[ 3\tan A - \tan^3 A = k(1 - 3\tan^2 A) \]

### Step 3: Rearranging the equation

Rearranging gives us:

\[ 3\tan A - \tan^3 A = k - 3k\tan^2 A \]

Rearranging further, we have:

\[ \tan^3 A - 3k\tan^2 A + 3\tan A - k = 0 \]

### Step 4: Let \(x = \tan A\)

Let \(x = \tan A\), then the equation becomes:

\[ x^3 - 3kx^2 + 3x - k = 0 \]

### Step 5: Find \(\frac{\sin 3A}{\sin A}\)

Using the sine triple angle formula:

\[ \sin 3A = 3\sin A - 4\sin^3 A \]

Thus,

\[ \frac{\sin 3A}{\sin A} = \frac{3\sin A - 4\sin^3 A}{\sin A} = 3 - 4\sin^2 A \]

### Step 6: Express \(\sin^2 A\) in terms of \(\tan^2 A\)

Since \(\sin^2 A = \frac{\tan^2 A}{1 + \tan^2 A}\), we can substitute:

\[ \sin^2 A = \frac{x^2}{1 + x^2} \]

### Step 7: Substitute \(\sin^2 A\) into the sine ratio

Substituting this into the sine ratio:

\[ \frac{\sin 3A}{\sin A} = 3 - 4\left(\frac{x^2}{1 + x^2}\right) = 3 - \frac{4x^2}{1 + x^2} \]

### Step 8: Find a common denominator

Combining the terms gives:

\[ \frac{\sin 3A}{\sin A} = \frac{3(1 + x^2) - 4x^2}{1 + x^2} = \frac{3 + 3x^2 - 4x^2}{1 + x^2} = \frac{3 - x^2}{1 + x^2} \]

### Step 9: Relate \(\tan^2 A\) to \(k\)

From earlier, we know:

\[ x^2 = \frac{k - 3}{3k - 1} \]

Substituting this back into the sine ratio:

\[ \frac{\sin 3A}{\sin A} = \frac{3 - \frac{k - 3}{3k - 1}}{1 + \frac{k - 3}{3k - 1}} = \frac{(3(3k - 1) - (k - 3))}{(3k - 1) + (k - 3)} \]

### Step 10: Simplify the expression

This simplifies to:

\[ \frac{(9k - 3 - k + 3)}{(3k - 1 + k - 3)} = \frac{(8k)}{(4k - 4)} = \frac{2k}{k - 1} \]

Thus, we have shown:

\[ \frac{\sin 3A}{\sin A} = \frac{2k}{k - 1} \]

### Step 11: Prove that \(k > 3\) or \(k < \frac{1}{3}\)

From the derived expression for \(\tan^2 A\):

\[ \tan^2 A = \frac{k - 3}{3k - 1} \]

For \(\tan^2 A\) to be non-negative, we need:

1. \(k - 3 \geq 0\) and \(3k - 1 > 0\) which gives \(k \geq 3\).
2. \(k - 3 < 0\) and \(3k - 1 < 0\) which gives \(k < \frac{1}{3}\).

Thus, we conclude:

\[ k > 3 \quad \text{or} \quad k < \frac{1}{3} \]