Answer
$$\frac{\tan^2t-1}{\sec^2t}=\frac{\tan t-\cot t}{\tan t+\cot t}$$
We simplify the left side to make it equal with the right side. It is an identity eventually.
Work Step by Step
$$\frac{\tan^2t-1}{\sec^2t}=\frac{\tan t-\cot t}{\tan t+\cot t}$$
We simplify the left side first.
$$A=\frac{\tan^2t-1}{\sec^2t}$$
- From a Pythagorean Identity: $$\sec^2t=1+\tan^2t$$
$$A=\frac{\tan^2 t-1}{1+\tan^2 t}$$
- We remember from a Reciprocal Identity: $$\cot t=\frac{1}{\tan t}$$
So, $$\cot t\tan t=1$$
That means we can replace $1$ with $\cot t\tan t$. In detail,
$$A=\frac{\tan^2 t-\tan t\cot t}{\tan^2 t+\tan t\cot t}$$
$$A=\frac{\tan t(\tan t-\cot t)}{\tan t(\tan t+\cot t)}$$
$$A=\frac{\tan t-\cot t}{\tan t+\cot t}$$
The left and right sides are equal. The expression is absolutely an identity.
