Let's prove the identity [tex]\((\sec^2(x) - 1) \csc^2(x) = \sec^2(x)\)[/tex].
### Step-by-Step Solution:
#### 1. Start with the left side (LS):
[tex]\[
\text{LS} = (\sec^2(x) - 1) \csc^2(x)
\][/tex]
#### 2. Utilize trigonometric identities:
- Recall that [tex]\(\sec^2(x) - 1 = \tan^2(x)\)[/tex].
- Also, recall that [tex]\(\csc^2(x) = 1 + \cot^2(x)\)[/tex].
#### 3. Substitute these identities into LS:
[tex]\[
\text{LS} = \tan^2(x) \times (1 + \cot^2(x))
\][/tex]
#### 4. Recall that [tex]\(\cot(x) = \frac{1}{\tan(x)}\)[/tex], so [tex]\(\cot^2(x) = \frac{1}{\tan^2(x)}\)[/tex]:
[tex]\[
\text{LS} = \tan^2(x) \times \left(1 + \frac{1}{\tan^2(x)}\right)
\][/tex]
#### 5. Simplify the expression inside the parentheses:
[tex]\[
1 + \frac{1}{\tan^2(x)} = \frac{\tan^2(x) + 1}{\tan^2(x)}
\][/tex]
#### 6. Plug this back into the equation:
[tex]\[
\text{LS} = \tan^2(x) \times \frac{\tan^2(x) + 1}{\tan^2(x)}
\][/tex]
#### 7. Simplify by canceling [tex]\(\tan^2(x)\)[/tex] from the numerator and the denominator:
[tex]\[
\text{LS} = \tan^2(x) + 1
\][/tex]
#### 8. Use the Pythagorean identity again:
[tex]\[
\tan^2(x) + 1 = \sec^2(x)
\][/tex]
#### 9. Thus, we have:
[tex]\[
\text{LS} = \sec^2(x)
\][/tex]
### Final Validation
Since [tex]\(\text{LS} = \sec^2(x)\)[/tex], and the right side (RS) of the original identity is also [tex]\(\sec^2(x)\)[/tex], we have:
[tex]\[
\text{LS} = \text{RS}
\][/tex]
Therefore, the given identity [tex]\((\sec^2(x) - 1) \csc^2(x) = \sec^2(x)\)[/tex] is proven to be true.
