To prove that:

$\left(\sec^2 x\right)\left(\csc^2 x\right) = \sec^2 x + \csc^2 x$

we will start by expressing $\sec^2 x$ and $\csc^2 x$ in terms of $\sin x$ and $\cos x$.

Step 1: Rewrite in terms of sine and cosine

Recall the definitions: $\sec^2 x = \frac{1}{\cos^2 x} \quad \text{and} \quad \csc^2 x = \frac{1}{\sin^2 x}$

Therefore, we can rewrite the left side of the equation as: $\left(\sec^2 x\right)\left(\csc^2 x\right) = \frac{1}{\cos^2 x} \cdot \frac{1}{\sin^2 x} = \frac{1}{\sin^2 x \cos^2 x}$

Step 2: Simplify using identities

Now, let's rewrite $\sec^2 x + \csc^2 x$ in terms of sine and cosine as well:

$\sec^2 x + \csc^2 x = \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x}$

To combine these, we need a common denominator, which is $\sin^2 x \cos^2 x$:

$\sec^2 x + \csc^2 x = \frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x}$

Step 3: Use the Pythagorean identity

From the Pythagorean identity, we know that: $\sin^2 x + \cos^2 x = 1$

Substituting this into our expression, we get: $\sec^2 x + \csc^2 x = \frac{1}{\sin^2 x \cos^2 x}$

Step 4: Conclusion

Now, we see that both sides of the equation are equal: $\left(\sec^2 x\right)\left(\csc^2 x\right) = \frac{1}{\sin^2 x \cos^2 x} = \sec^2 x + \csc^2 x$

Therefore, we have proven that: $\left(\sec^2 x\right)\left(\csc^2 x\right) = \sec^2 x + \csc^2 x$

Would you like further details on any part of this proof?

1. Why is $\sin^2 x + \cos^2 x = 1$ important in trigonometric proofs?
2. How can we prove other trigonometric identities using similar methods?
3. What are the common applications of this identity in calculus?
4. Can we verify this identity with specific values of $x$?
5. How can we extend this proof to other trigonometric identities?

Tip: Expressing trigonometric functions in terms of sine and cosine often simplifies complex proofs.