We are tasked with simplifying or proving the identity:

$\cot\theta + \tan\theta = \csc\theta \cdot \sec\theta$

Step 1: Rewrite all terms in terms of $\sin\theta$ and $\cos\theta$

Using the following trigonometric definitions:

• $\cot\theta = \frac{\cos\theta}{\sin\theta}$
• $\tan\theta = \frac{\sin\theta}{\cos\theta}$
• $\csc\theta = \frac{1}{\sin\theta}$
• $\sec\theta = \frac{1}{\cos\theta}$

The equation becomes: $\frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta} = \frac{1}{\sin\theta} \cdot \frac{1}{\cos\theta}.$

Step 2: Simplify the left-hand side (LHS)

The LHS is: $\frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta}.$

Finding a common denominator ($\sin\theta \cos\theta$): $\frac{\cos^2\theta + \sin^2\theta}{\sin\theta \cos\theta}.$

Using the Pythagorean identity $\cos^2\theta + \sin^2\theta = 1$, this simplifies to: $\frac{1}{\sin\theta \cos\theta}.$

Step 3: Simplify the right-hand side (RHS)

The RHS is: $\frac{1}{\sin\theta} \cdot \frac{1}{\cos\theta}.$

This simplifies to: $\frac{1}{\sin\theta \cos\theta}.$

Step 4: Compare LHS and RHS

Both sides are equal: $\frac{1}{\sin\theta \cos\theta} = \frac{1}{\sin\theta \cos\theta}.$

Thus, the identity is proven: $\boxed{\cot\theta + \tan\theta = \csc\theta \cdot \sec\theta}.$

---

Do you have any questions or need more details about this proof?

Related Questions:

1. What is the difference between $\cot\theta$ and $\tan\theta$?
2. How can Pythagorean identities help simplify trigonometric equations?
3. What are the domain restrictions for $\cot\theta$ and $\tan\theta$?
4. Can this identity be derived graphically or geometrically?
5. How do reciprocal trigonometric functions relate to their primary counterparts?

Tip:

Always look for opportunities to rewrite trigonometric expressions in terms of $\sin\theta$ and (\cos\theta