(5) \( \frac{\tan ^{2} \theta}{\tan ^{2} \theta+1}=\sin ^{2} \theta \)

To solve the equation \( \frac{\tan^2 \theta}{\tan^2 \theta + 1} = \sin^2 \theta \), we can start by rewriting \(\tan^2 \theta\) in terms of \(\sin\) and \(\cos\): \[ \tan^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} \] Substituting this into the equation gives: \[ \frac{\frac{\sin^2 \theta}{\cos^2 \theta}}{\frac{\sin^2 \theta}{\cos^2 \theta} + 1} = \sin^2 \theta \] Now, simplifying the denominator: \[ \frac{\sin^2 \theta}{\cos^2 \theta} + 1 = \frac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta} \] Then, we can rewrite our main equation: \[ \frac{\sin^2 \theta}{\frac{1}{\cos^2 \theta}} = \sin^2 \theta \] This simplifies to: \[ \sin^2 \theta \cos^2 \theta = \sin^2 \theta \] Assuming \(\sin^2 \theta \neq 0\) (which we can check later), we can divide both sides by \(\sin^2 \theta\): \[ \cos^2 \theta = 1 \] This implies: \[ \cos \theta = \pm 1 \] Which means \(\theta = n\pi\), where \(n\) is any integer. If we had considered the case \(\sin^2 \theta = 0\), we would also find \(\theta = k\pi\) for \(k\) any integer, coinciding with our previous solution. Thus, the final solution is: \[ \theta = n\pi \quad (n \text{ is an integer}). \]