We are given the following trigonometric identity:

$\frac{\tan(78^\circ) - \tan(18^\circ)}{1 + \tan(78^\circ) \cdot \tan(18^\circ)} = \tan(A^\circ) = \sqrt{B}.$

Step 1: Use the tangent subtraction identity.

This is an application of the tangent subtraction formula:

$\tan(x - y) = \frac{\tan(x) - \tan(y)}{1 + \tan(x) \cdot \tan(y)}.$

If we compare this with the given equation, we can rewrite it as:

$\frac{\tan(78^\circ) - \tan(18^\circ)}{1 + \tan(78^\circ) \cdot \tan(18^\circ)} = \tan((78^\circ - 18^\circ)) = \tan(60^\circ).$

Thus, $A = 60^\circ$.

Step 2: Solve for $B$.

Since we now know that $\tan(A^\circ) = \tan(60^\circ)$, and $\tan(60^\circ) = \sqrt{3}$, we have:

$\sqrt{B} = \sqrt{3}.$

By squaring both sides:

$B = 3.$

Final Answer:

• $A = 60^\circ$
• $B = 3$

Would you like more details on any part of this solution?

Here are some related questions:

1. How does the tangent subtraction formula work in other cases?
2. Why does $\tan(60^\circ) = \sqrt{3}$?
3. What are some other important trigonometric identities involving tangent?
4. Can we derive this result using a different approach?
5. What happens if we change the angles in the original equation?

Tip: Tangent identities like these often simplify complex angle relationships, especially when using known angles such as 60° or 45°.