Answer
$$\frac{\cot^2 t-1}{1+\cot^2t}=1-2\sin^2t$$
The given trigonometric expression is an identity.
Work Step by Step
$$\frac{\cot^2 t-1}{1+\cot^2t}=1-2\sin^2t$$
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.
