## Trigonometry (11th Edition) Clone

$$\frac{\cot^2 t-1}{1+\cot^2t}=1-2\sin^2t$$ The given trigonometric expression is an identity.
$$\frac{\cot^2 t-1}{1+\cot^2t}=1-2\sin^2t$$ 1) We simplify the left side. $$A=\frac{\cot^2t-1}{1+\cot^2t}$$ - Using the Quotient Identity: $$\cot t=\frac{\cos t}{\sin t}$$ Therefore, $$A=\frac{\frac{\cos^2t}{\sin^2 t}-1}{1+\frac{\cos^2t}{\sin^2t}}$$ $$A=\frac{\frac{\cos^2t-\sin^2t}{\sin^2 t}}{\frac{\sin^2t+\cos^2t}{\sin^2t}}$$ $$A=\frac{\cos^2t-\sin^2t}{\sin^2 t}\times\frac{\sin^2 t}{\sin^2t+\cos^2t}$$ $$A=\frac{\cos^2t-\sin^2t}{\sin^2t+\cos^2t}$$ - Using the Pythagorean Identity: $$\sin^2t+\cos^2t=1$$ Therefore, $$A=\frac{\cos^2t-\sin^2t}{1}$$ $$A=\cos^2t-\sin^2t$$ 2) We now look at the right side: $$B=1-2\sin^2 t$$ - Remember the Pythagorean Identity: $$\sin^2t+\cos^2t=1$$ Replace into $B$: $$B=\sin^2t+\cos^2t-2\sin^2t$$ $$B=\cos^2t-\sin^2t$$ 3) Therefore, $A$ is equal with $B$. The expression is an identity.