`tan75^@+tan15^@=….`

Options

A. `1+sqrt3`
B. `sqrt3-1`
C. `sqrt3`
D. 4

Answer

To find the value of \( \tan 75^\circ + \tan 15^\circ \), we can use the tangent addition formula. The tangent addition formula states that:

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

In this case, we can express \( 75^\circ \) and \( 15^\circ \) in terms of \( 45^\circ \) and \( 30^\circ \):

\[ \tan 75^\circ = \tan(45^\circ + 30^\circ) \]
\[ \tan 15^\circ = \tan(45^\circ - 30^\circ) \]

Using the addition and subtraction formulas:

1. **Calculate \( \tan 75^\circ \)**:
\[ \tan 75^\circ = \tan(45^\circ + 30^\circ) = \frac{\tan 45^\circ + \tan 30^\circ}{1 - \tan 45^\circ \tan 30^\circ} \]
Where \( \tan 45^\circ = 1 \) and \( \tan 30^\circ = \frac{1}{\sqrt{3}} \).

Substituting these values:
\[ \tan 75^\circ = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} \]

2. **Calculate \( \tan 15^\circ \)**:
\[ \tan 15^\circ = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} \]
Substituting the values:
\[ \tan 15^\circ = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} \]

3. **Combine \( \tan 75^\circ \) and \( \tan 15^\circ \)**:
Now we need to add these two results:
\[ \tan 75^\circ + \tan 15^\circ = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} + \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} \]

4. **Finding a common denominator**:
The common denominator for the two fractions is \( (1 - \frac{1}{\sqrt{3}})(1 + \frac{1}{\sqrt{3}}) \).

Simplifying:
\[ \tan 75^\circ + \tan 15^\circ = \frac{(1 + \frac{1}{\sqrt{3}})^2 + (1 - \frac{1}{\sqrt{3}})^2}{(1 - \frac{1}{\sqrt{3}})(1 + \frac{1}{\sqrt{3}})} \]

5. **Calculating the numerator**:
Expanding the numerator:
\[ (1 + \frac{1}{\sqrt{3}})^2 = 1 + 2 \cdot \frac{1}{\sqrt{3}} + \frac{1}{3} = \frac{4}{3} + \frac{2}{\sqrt{3}} \]
\[ (1 - \frac{1}{\sqrt{3}})^2 = 1 - 2 \cdot \frac{1}{\sqrt{3}} + \frac{1}{3} = \frac{4}{3} - \frac{2}{\sqrt{3}} \]

Adding these:
\[ \frac{4}{3} + \frac{2}{\sqrt{3}} + \frac{4}{3} - \frac{2}{\sqrt{3}} = \frac{8}{3} \]

6. **Calculating the denominator**:
The denominator simplifies to:
\[ (1 - \frac{1}{\sqrt{3}})(1 + \frac{1}{\sqrt{3}}) = 1 - \frac{1}{3} = \frac{2}{3} \]

7. **Final calculation**:
\[ \tan 75^\circ + \tan 15^\circ = \frac{\frac{8}{3}}{\frac{2}{3}} = 4 \]

Thus, the final answer is:

\[ \boxed{4} \]