## Trigonometry (11th Edition) Clone

$$\frac{\tan^2t-1}{\sec^2t}=\frac{\tan t-\cot t}{\tan t+\cot t}$$ We simplify the left side to make it equal with the right side. It is an identity eventually.
$$\frac{\tan^2t-1}{\sec^2t}=\frac{\tan t-\cot t}{\tan t+\cot t}$$ 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.