Answer
$$\frac{\sec^2\theta-1}{\csc^2\theta-1}=\tan^4\theta$$
Work Step by Step
$$A=\frac{\sec^2\theta-1}{\csc^2\theta-1}$$
- Reciprocal Identities:
$$\sec\theta=\frac{1}{\cos\theta}$$
$$\csc\theta=\frac{1}{\sin\theta}$$
Replace into $A$:
$$A=\frac{\frac{1}{\cos^2\theta}-1}{\frac{1}{\sin^2\theta}-1}$$
$$A=\frac{\frac{1-\cos^2\theta}{\cos^2\theta}}{\frac{1-\sin^2\theta}{\sin^2\theta}}$$
- Pythagorean Identities:
$$1-\cos^2\theta=\sin^2\theta$$
$$1-\sin^2\theta=\cos^2\theta$$
That means
$$A=\frac{\frac{\sin^2\theta}{\cos^2\theta}}{\frac{\cos^2\theta}{\sin^2\theta}}$$
$$A=\frac{\sin^2\theta\times\sin^2\theta}{\cos^2\theta\times\cos^2\theta}$$
$$A=\frac{\sin^4\theta}{\cos^4\theta}$$
$$A=\Bigg(\frac{\sin\theta}{\cos\theta}\Bigg)^4$$
- Quotient Identity:
$$\tan\theta=\frac{\sin\theta}{\cos\theta}$$
Replace into $A$:
$$A=\tan^4\theta$$
