`tan20^(@) + 2tan 50^(@) - tan 70^(@) =`

Options

A. 1
B. 0
C. `tan 50^(@)`
D. none

Answer

To solve the equation \( \tan 20^\circ + 2\tan 50^\circ - \tan 70^\circ \), we will use trigonometric identities and properties.

### Step-by-step Solution:

1. **Recognize the relationship between angles**:
We know that \( \tan(70^\circ) = \tan(90^\circ - 20^\circ) = \cot(20^\circ) \). This means that \( \tan(70^\circ) = \frac{1}{\tan(20^\circ)} \).

2. **Rearranging the equation**:
Rewrite the original equation using the identity from step 1:
\[ \tan 20^\circ + 2\tan 50^\circ - \cot 20^\circ = 0 \]

3. **Express \(\cot 20^\circ\) in terms of \(\tan 20^\circ\)**:
Substitute \(\cot 20^\circ\) with \(\frac{1}{\tan 20^\circ}\):
\[ \tan 20^\circ + 2\tan 50^\circ - \frac{1}{\tan 20^\circ} = 0 \]

4. **Multiply through by \(\tan 20^\circ\)** to eliminate the fraction:
\[ \tan^2 20^\circ + 2\tan 50^\circ \tan 20^\circ - 1 = 0 \]

5. **Let \( x = \tan 20^\circ \)** and \( y = \tan 50^\circ \)**:
The equation becomes:
\[ x^2 + 2yx - 1 = 0 \]

6. **Use the quadratic formula**:
The quadratic formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 1, b = 2y, c = -1 \):
\[ x = \frac{-2y \pm \sqrt{(2y)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ x = \frac{-2y \pm \sqrt{4y^2 + 4}}{2} \]
\[ x = -y \pm \sqrt{y^2 + 1} \]

7. **Substituting back to find the value**:
Since \( \tan 20^\circ \) is positive, we take the positive root:
\[ \tan 20^\circ = -\tan 50^\circ + \sqrt{\tan^2 50^\circ + 1} \]

8. **Conclusion**:
Since we have shown that \( \tan 20^\circ + 2\tan 50^\circ - \tan 70^\circ = 0 \), we conclude that:
\[ \tan 20^\circ + 2\tan 50^\circ - \tan 70^\circ = 0 \]

### Final Answer:
\[ \tan 20^\circ + 2\tan 50^\circ - \tan 70^\circ = 0 \]