To find the exact value of $\tan(75^\circ)$, we can use the sum identity for tangent. We know that:

$75^\circ = 45^\circ + 30^\circ$

Using the sum identity for tangent:

$\tan(A + B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)}$

Let $A = 45^\circ$ and $B = 30^\circ$, so:

$\tan(75^\circ) = \frac{\tan(45^\circ) + \tan(30^\circ)}{1 - \tan(45^\circ)\tan(30^\circ)}$

We know the exact values of $\tan(45^\circ)$ and $\tan(30^\circ)$:

• $\tan(45^\circ) = 1$
• $\tan(30^\circ) = \frac{1}{\sqrt{3}}$

Now substitute these values into the identity:

$\tan(75^\circ) = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}}$

Simplify the numerator and the denominator:

$\tan(75^\circ) = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}}$

Now, to simplify further, let's make both the numerator and denominator have a common denominator:

Numerator: $1 + \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{\sqrt{3} + 1}{\sqrt{3}}$

Denominator: $1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}}$

Now the expression becomes:

$\tan(75^\circ) = \frac{\frac{\sqrt{3} + 1}{\sqrt{3}}}{\frac{\sqrt{3} - 1}{\sqrt{3}}}$

The $\sqrt{3}$ terms cancel out, leaving:

$\tan(75^\circ) = \frac{\sqrt{3} + 1}{\sqrt{3} - 1}$

This is the exact value of $\tan(75^\circ)$.

Would you like any further simplifications or clarifications?

Related questions:

1. How can we simplify further expressions involving trigonometric sums?
2. What are other angles that use sum and difference identities?
3. Can the tangent of non-standard angles be found without using a calculator?
4. How do half-angle identities differ from sum and difference identities?
5. What are other methods to find exact values for trigonometric functions?

Tip: Learning the exact values for common angles (like 30°, 45°, and 60°) helps greatly in simplifying trigonometric problems without a calculator.