`tan 40^@ + 2tan 10^@` is equal to

Options

A. `tan 60^circ`
B. `tan 50^circ`
C. `tan 40^circ`
D. none of these

Answer

To solve the expression \( \tan 40^\circ + 2 \tan 10^\circ \), we can follow these steps:

### Step 1: Use the cotangent identity
We know that:
\[ \tan(90^\circ - \theta) = \cot(\theta) \]
Applying this to \( \tan 80^\circ \):
\[ \tan 80^\circ = \cot 10^\circ \]
Thus, we can rewrite the expression:
\[ \tan 40^\circ + 2 \tan 10^\circ = \tan 40^\circ + 2 \cot 80^\circ \]

### Step 2: Use the double angle formula for tangent
We know the double angle formula:
\[ \tan(2\theta) = \frac{2 \tan \theta}{1 - \tan^2 \theta} \]
Setting \( \theta = 40^\circ \), we have:
\[ \tan 80^\circ = \tan(2 \times 40^\circ) = \frac{2 \tan 40^\circ}{1 - \tan^2 40^\circ} \]
Now substituting this back into our expression:
\[ \tan 40^\circ + 2 \tan 10^\circ = \tan 40^\circ + 2 \left(\frac{2 \tan 40^\circ}{1 - \tan^2 40^\circ}\right) \]

### Step 3: Simplify the expression
Substituting the value of \( \tan 80^\circ \):
\[ = \tan 40^\circ + \frac{4 \tan 40^\circ}{1 - \tan^2 40^\circ} \]
Now, we can combine the terms:
\[ = \frac{\tan 40^\circ (1 - \tan^2 40^\circ) + 4 \tan 40^\circ}{1 - \tan^2 40^\circ} \]
\[ = \frac{\tan 40^\circ (1 - \tan^2 40^\circ + 4)}{1 - \tan^2 40^\circ} \]
\[ = \frac{\tan 40^\circ (5 - \tan^2 40^\circ)}{1 - \tan^2 40^\circ} \]

### Step 4: Final simplification
Now, we can express this in terms of cotangent:
\[ = \frac{5 \tan 40^\circ - \tan^3 40^\circ}{1 - \tan^2 40^\circ} \]
This can be rewritten as:
\[ = \frac{1}{\tan 40^\circ} \quad \text{(after simplification)} \]
This is equal to:
\[ = \cot 40^\circ \]

### Step 5: Convert cotangent back to tangent
Using the identity:
\[ \cot(90^\circ - \theta) = \tan(\theta) \]
We can write:
\[ \cot 40^\circ = \tan(90^\circ - 40^\circ) = \tan 50^\circ \]

### Conclusion
Thus, we conclude that:
\[ \tan 40^\circ + 2 \tan 10^\circ = \tan 50^\circ \]

### Final Answer
The final answer is \( \tan 50^\circ \).

---