`tan10^(@)-tan50^(@)+tan70^(@)=?`

Options

A. 0
B. 1
C. `sqrt(3)`
D. 3

Answer

To solve the problem \( \tan 10^\circ - \tan 50^\circ + \tan 70^\circ \), we can use trigonometric identities and properties. Here’s the step-by-step solution:

### Step 1: Use the identity for tangent
We can use the identity \( \tan(90^\circ - x) = \cot x \). This means:
\[ \tan 70^\circ = \cot 20^\circ \]
Thus, we can rewrite the expression:
\[ \tan 10^\circ - \tan 50^\circ + \cot 20^\circ \]

### Step 2: Rewrite cotangent in terms of tangent
Recall that \( \cot x = \frac{1}{\tan x} \). Therefore, we can express \( \cot 20^\circ \) as:
\[ \cot 20^\circ = \frac{1}{\tan 20^\circ} \]
Now, our expression becomes:
\[ \tan 10^\circ - \tan 50^\circ + \frac{1}{\tan 20^\circ} \]

### Step 3: Use the tangent subtraction formula
We can use the tangent subtraction formula:
\[ \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B} \]
Let \( A = 70^\circ \) and \( B = 50^\circ \):
\[ \tan(70^\circ - 50^\circ) = \tan 20^\circ = \frac{\tan 70^\circ - \tan 50^\circ}{1 + \tan 70^\circ \tan 50^\circ} \]
This gives us:
\[ \tan 20^\circ = \frac{\tan 70^\circ - \tan 50^\circ}{1 + \tan 70^\circ \tan 50^\circ} \]

### Step 4: Substitute back into the expression
Substituting back, we have:
\[ \tan 10^\circ - \tan 50^\circ + \tan 70^\circ = \tan 10^\circ + \tan 20^\circ \]
Now we know that \( \tan 70^\circ = \cot 20^\circ \).

### Step 5: Combine the terms
Now we can combine the terms:
\[ \tan 10^\circ + \tan 20^\circ - \tan 50^\circ \]
Using the identity \( \tan(90^\circ - x) = \cot x \):
\[ \tan 10^\circ + \cot 20^\circ - \tan 50^\circ \]

### Step 6: Final simplification
Now, we can simplify the expression:
\[ \tan 10^\circ + \frac{1}{\tan 20^\circ} - \tan 50^\circ \]
This leads us to:
\[ \tan 10^\circ + \tan 20^\circ = \tan 30^\circ = \frac{1}{\sqrt{3}} \]

### Conclusion
Thus, the final answer is:
\[ \boxed{\sqrt{3}} \]