We are given the trigonometric equation:

$\csc^2x \cdot \tan^2x - 1 = \tan^2x$

Let's solve this step by step.

Step 1: Recall the identity for $\csc^2x$

We know that:

$\csc^2x = 1 + \cot^2x$

This identity will help simplify the expression. Substituting $\csc^2x$ in the given equation:

$(1 + \cot^2x) \cdot \tan^2x - 1 = \tan^2x$

Step 2: Expand the left-hand side

Expand the left-hand side of the equation:

$\tan^2x + \cot^2x \cdot \tan^2x - 1 = \tan^2x$

Step 3: Simplify $\cot^2x \cdot \tan^2x$

Since $\cot x = \frac{1}{\tan x}$, we have:

$\cot^2x \cdot \tan^2x = 1$

So the equation becomes:

$\tan^2x + 1 - 1 = \tan^2x$

Step 4: Simplify the equation

The terms $+1$ and $-1$ cancel each other out:

$\tan^2x = \tan^2x$

This is an identity, which means the equation is true for all values of $x$ where the trigonometric functions are defined. Therefore, the equation holds for all values of $x$ except those that would make $\tan x$ or $\csc x$ undefined.

Final Answer:

The equation $\csc^2x \cdot \tan^2x - 1 = \tan^2x$ is an identity and holds for all $x$ where $\tan x$ and $\csc x$ are defined.

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Do you have any questions or would you like further details on any step? Here are some related questions you might consider:

1. What values of $x$ make $\tan x$ and $\csc x$ undefined?
2. Can you prove other trigonometric identities involving $\tan x$ and $\csc x$?
3. How do the Pythagorean identities relate to this equation?
4. What are the implications of this being an identity rather than a conditional equation?
5. How does this identity change when considering radians vs. degrees?

Tip: Always check the domain of trigonometric functions to ensure solutions are valid within the range.