Factor the trigonometric expression using the identity \( 1 - \tan^2(x) = \frac{1}{\sec^2(x)} \): \( 1 - \sin^2(x)\tan^2(x) \).

To factor the expression \( 1 - \sin^2(x)\tan^2(x) \), we can start by using the identity \( \tan(x) = \frac{\sin(x)}{\cos(x)} \) to rewrite \(\tan^2(x)\): \[ \tan^2(x) = \frac{\sin^2(x)}{\cos^2(x)}. \] Substituting \(\tan^2(x)\) into the expression gives: \[ 1 - \sin^2(x)\tan^2(x) = 1 - \sin^2(x) \cdot \frac{\sin^2(x)}{\cos^2(x)} = 1 - \frac{\sin^4(x)}{\cos^2(x)}. \] Next, we can factor out \( \frac{1}{\cos^2(x)} \): \[ 1 - \frac{\sin^4(x)}{\cos^2(x)} = \frac{1}{\cos^2(x)} \left( \cos^2(x) - \sin^4(x) \right). \] Now we can further factor \( \cos^2(x) - \sin^4(x) \) using the difference of squares: \[ \cos^2(x) - \sin^4(x) = \cos^2(x) - (\sin^2(x))^2 = (\cos(x) - \sin^2(x))(\cos(x) + \sin^2(x)). \] Putting it all together, we get: \[ 1 - \sin^2(x)\tan^2(x) = \frac{1}{\cos^2(x)} \left( (\cos(x) - \sin^2(x))(\cos(x) + \sin^2(x)) \right). \] Thus, the final factored form is: \[ 1 - \sin^2(x)\tan^2(x) = \frac{(\cos(x) - \sin^2(x))(\cos(x) + \sin^2(x))}{\cos^2(x)}. \]