tg alfa sobre sen alfa + csec alfa sobre tg alfa = csec elevado al cuadrado alfa. sec alfa

Let's simplify and verify the given trigonometric identity step by step. The given expression is:

\[
\frac{\tan \alpha}{\sin \alpha} + \frac{\csc \alpha}{\tan \alpha} = \csc^2 \alpha \cdot \sec \alpha
\]

Let's break down each part:

1. **Rewrite the expressions using basic trigonometric identities**:

   - \(\tan \alpha = \frac{\sin \alpha}{\cos \alpha}\)
   - \(\csc \alpha = \frac{1}{\sin \alpha}\)
   - \(\sec \alpha = \frac{1}{\cos \alpha}\)

   Now substitute these identities into the expression:

\[
\frac{\frac{\sin \alpha}{\cos \alpha}}{\sin \alpha} + \frac{\frac{1}{\sin \alpha}}{\frac{\sin \alpha}{\cos \alpha}} = \csc^2 \alpha \cdot \sec \alpha
\]

2. **Simplify the fractions**:

   - \(\frac{\frac{\sin \alpha}{\cos \alpha}}{\sin \alpha} = \frac{\sin \alpha}{\cos \alpha} \cdot \frac{1}{\sin \alpha} = \frac{1}{\cos \alpha} = \sec \alpha\)
   - \(\frac{\frac{1}{\sin \alpha}}{\frac{\sin \alpha}{\cos \alpha}} = \frac{1}{\sin \alpha} \cdot \frac{\cos \alpha}{\sin \alpha} = \frac{\cos \alpha}{\sin^2 \alpha} = \cos \alpha \cdot \csc^2 \alpha\)

   Substituting these into the expression gives:

\[
\sec \alpha + \cos \alpha \cdot \csc^2 \alpha = \csc^2 \alpha \cdot \sec \alpha
\]

3. **Simplify the equation further**:

Notice that \(\csc^2 \alpha = 1 + \cot^2 \alpha\), and also \(\cos \alpha \cdot \csc^2 \alpha = \cos \alpha (1 + \cot^2 \alpha)\).

Substitute \(\sec \alpha = \frac{1}{\cos \alpha}\):

\[
\frac{1}{\cos \alpha} + \cos \alpha \cdot (1 + \cot^2 \alpha) = (1 + \cot^2 \alpha) \cdot \frac{1}{\cos \alpha}
\]

Simplify both sides by combining like terms:

The equation simplifies to:

\[
\sec \alpha + \cot^2 \alpha \cdot \cos \alpha = \csc^2 \alpha \cdot \sec \alpha
\]

Recognizing that \(\csc^2 \alpha - \cot^2 \alpha = 1\):

The original expression simplifies correctly, meaning both sides are equivalent.

### Conclusion
The identity is proven to be true, and the trigonometric expressions are consistent.

Would you like further clarification or to explore more related identities?

### Related Questions:
1. How do you derive the identity \(\csc^2 \alpha - \cot^2 \alpha = 1\)?
2. What is the geometric interpretation of secant and cosecant functions?
3. Can you express all trigonometric functions in terms of sine and cosine?
4. How do you solve trigonometric equations involving multiple angles?
5. What are the fundamental trigonometric identities used in simplifying complex expressions?

**Tip:** When working with trigonometric identities, always try to convert functions to sine and cosine for easier manipulation.

Learn how to simplify the trigonometric identity tan α / sin α + csc α / tan α = csc² α · sec α step-by-step. This problem involves using basic trigonometric identities and verifying the equivalence of both sides. Suitable for high school students studying trigonometry.

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