1. **State the problem:** Simplify the expression $$\frac{\tan^2 t - 1}{\sec^2 t} = \frac{\tan t - \cot t}{\tan t + \cot t}$$ 2. **Recall trigonometric identities:** - $\tan^2 t + 1 = \sec^2 t$ - $\cot t = \frac{1}{\tan t}$ 3. **Simplify the left side:** Using $\tan^2 t - 1 = (\tan^2 t + 1) - 2 = \sec^2 t - 2$ is incorrect, so instead use the identity $\tan^2 t + 1 = \sec^2 t$ to rewrite numerator: $$\tan^2 t - 1 = (\sec^2 t - 1) - 1 = \sec^2 t - 2$$ But this is not a standard identity, so better to keep as is and divide numerator and denominator: $$\frac{\tan^2 t - 1}{\sec^2 t} = \frac{\tan^2 t - 1}{\tan^2 t + 1}$$ since $\sec^2 t = 1 + \tan^2 t$. 4. **Simplify the right side:** Rewrite $\cot t$ as $\frac{1}{\tan t}$: $$\frac{\tan t - \frac{1}{\tan t}}{\tan t + \frac{1}{\tan t}} = \frac{\frac{\tan^2 t - 1}{\tan t}}{\frac{\tan^2 t + 1}{\tan t}}$$ 5. **Cancel common factor $\tan t$ in numerator and denominator:** $$= \frac{\tan^2 t - 1}{\tan^2 t + 1}$$ 6. **Conclusion:** Both sides simplify to the same expression: $$\frac{\tan^2 t - 1}{\tan^2 t + 1}$$ Therefore, the original equation holds true. **Final simplified form:** $$\boxed{\frac{\tan^2 t - 1}{\tan^2 t + 1}}$$