Prove(trigonometry):

tan(x)tan(60-x)tan(60+x)=tan(3x)

To prove the identity \( \tan(x)\tan(60^\circ - x)\tan(60^\circ + x) = \tan(3x) \), we can use trigonometric identities and properties of tangent. This identity involves manipulation of angles and employing the tangent addition and subtraction formulas. Let's break it down step by step.

Step 1: Understand the Components

First, we recognize the angles involved. The angle \(60^\circ\) has specific values for sine and cosine that can be useful:

• \( \tan(60^\circ) = \sqrt{3} \)
• \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \)
• \( \cos(60^\circ) = \frac{1}{2} \)

We will also need the tangent addition and subtraction formulas:

• \( \tan(a \pm b) = \frac{\tan a \pm \tan b}{1 \mp \tan a \tan b} \)

Step 2: Expand the Left-Hand Side

We start by expanding \( \tan(60^\circ - x) \) and \( \tan(60^\circ + x) \):

• \( \tan(60^\circ - x) = \frac{\tan(60^\circ) - \tan(x)}{1 + \tan(60^\circ) \tan(x)} = \frac{\sqrt{3} - \tan(x)}{1 + \sqrt{3}\tan(x)} \)
• \( \tan(60^\circ + x) = \frac{\tan(60^\circ) + \tan(x)}{1 - \tan(60^\circ) \tan(x)} = \frac{\sqrt{3} + \tan(x)}{1 - \sqrt{3}\tan(x)} \)

Step 3: Substitute and Simplify

Now we substitute these into the left-hand side of the original equation:

• Let \( T = \tan(x) \)
• Then, the left side becomes \( T \cdot \left(\frac{\sqrt{3} - T}{1 + \sqrt{3}T}\right) \cdot \left(\frac{\sqrt{3} + T}{1 - \sqrt{3}T}\right) \)

This simplifies to:

= \frac{T \cdot (\sqrt{3} - T)(\sqrt{3} + T)}{(1 + \sqrt{3}T)(1 - \sqrt{3}T)}= \frac{T(3 - T^2)}{1 - 3T^2}

Step 4: Simplify Further

We can express \( \tan(3x) \) using the triple angle formula:

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

Substituting \( T \) for \( \tan(x) \):

= \frac{3T - T^3}{1 - 3T^2}

Step 5: Equate Both Sides

Now we have two expressions:

• Left-hand side: \( \frac{T(3 - T^2)}{1 - 3T^2} \)
• Right-hand side: \( \frac{3T - T^3}{1 - 3T^2} \)

We see that both sides are equal:

T(3 - T^2) = 3T - T^3

This holds true, confirming the identity.

Final Thoughts

Thus, we've successfully proven the identity \( \tan(x)\tan(60^\circ - x)\tan(60^\circ + x) = \tan(3x) \) using fundamental trigonometric identities and properties. This approach not only demonstrates the validity of the identity but also reinforces the relationships between various angles and their tangential values.