Answer
$$\frac{1+\tan^2\theta}{1+\cot^2\theta}=\tan^2\theta$$
Work Step by Step
$$A=\frac{1+\tan^2\theta}{1+\cot^2\theta}$$
- Quotient Identities:
$$\tan\theta=\frac{\sin\theta}{\cos\theta}$$
$$\cot\theta=\frac{\cos\theta}{\sin\theta}$$
Replace into $A$:
$$A=\frac{1+\frac{\sin^2\theta}{\cos^2\theta}}{1+\frac{\cos^2\theta}{\sin^2\theta}}$$
$$A=\frac{\frac{\cos^2\theta+\sin^2\theta}{\cos^2\theta}}{\frac{\sin^2\theta+\cos^2\theta}{\sin^2\theta}}$$
$$A=\frac{(\cos^2\theta+\sin^2\theta)\sin^2\theta}{\cos^2\theta(\cos^2\theta+\sin^2\theta)}$$
$$A=\frac{\sin^2\theta}{\cos^2\theta}$$
$$A=\Bigg(\frac{\sin\theta}{\cos\theta}\Bigg)^2$$
- Quotient Identity again:
$$\tan\theta=\frac{\sin\theta}{\cos\theta}$$
Replace into $A$:
$$A=\tan^2\theta$$