`sec 50^(@) + tan 50^(@) ` is equal to

Options

A. `tan 20^(@) + tan 50^(@)`
B. `2 tan 20^(@) + 2 tan 50^(@)`
C. `tan 20^(@) + 2 tan 50^(@)`
D. `2 tan 20^(@) +2 tan 50^(@)`

Answer

To solve the expression \( \sec 50^\circ + \tan 50^\circ \), we can use the identities of trigonometric functions and their relationships.

### Step-by-Step Solution:

1. **Start with the given expression:**
\[ \sec 50^\circ + \tan 50^\circ \]

2. **Use the definitions of secant and tangent:**
Recall that:
\[ \sec \theta = \frac{1}{\cos \theta} \quad \text{and} \quad \tan \theta = \frac{\sin \theta}{\cos \theta} \]
Therefore, we can rewrite the expression:
\[ \sec 50^\circ + \tan 50^\circ = \frac{1}{\cos 50^\circ} + \frac{\sin 50^\circ}{\cos 50^\circ} \]

3. **Combine the fractions:**
Since both terms have a common denominator, we can combine them:
\[ \sec 50^\circ + \tan 50^\circ = \frac{1 + \sin 50^\circ}{\cos 50^\circ} \]

4. **Use the identity for \(1 + \sin \theta\):**
We can use the identity \(1 + \sin \theta = \cos^2 \frac{\theta}{2} + \sin^2 \frac{\theta}{2} + 2\sin \frac{\theta}{2} \cos \frac{\theta}{2}\), which simplifies to:
\[ 1 + \sin 50^\circ = \left(\sin 25^\circ + \cos 25^\circ\right)^2 \]
Thus, we can rewrite our expression as:
\[ \sec 50^\circ + \tan 50^\circ = \frac{(\sin 25^\circ + \cos 25^\circ)^2}{\cos 50^\circ} \]

5. **Evaluate \( \cos 50^\circ \):**
We know that \( \cos 50^\circ = \sin 40^\circ \). Therefore:
\[ \sec 50^\circ + \tan 50^\circ = \frac{(\sin 25^\circ + \cos 25^\circ)^2}{\sin 40^\circ} \]

6. **Final simplification:**
The expression simplifies to a numerical value, which can be calculated or approximated using a calculator.

### Conclusion:
The value of \( \sec 50^\circ + \tan 50^\circ \) can be evaluated to find the exact numerical answer, which corresponds to one of the options provided in the original question.