(6) \( \sin ^{2} \theta+\sin ^{2} \theta \tan ^{2} \theta=\tan ^{2} \theta \)

To solve the equation \( \sin^{2} \theta + \sin^{2} \theta \tan^{2} \theta = \tan^{2} \theta \), we start by recalling that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Therefore, \( \tan^{2} \theta = \frac{\sin^{2} \theta}{\cos^{2} \theta} \). Substituting this in, we get: \[ \sin^{2} \theta + \sin^{2} \theta \left( \frac{\sin^{2} \theta}{\cos^{2} \theta} \right) = \frac{\sin^{2} \theta}{\cos^{2} \theta} \] This simplifies to: \[ \sin^{2} \theta + \frac{\sin^{4} \theta}{\cos^{2} \theta} = \frac{\sin^{2} \theta}{\cos^{2} \theta} \] Multiplying through by \( \cos^{2} \theta \) (assuming \( \cos \theta \neq 0 \)) gives: \[ \sin^{2} \theta \cos^{2} \theta + \sin^{4} \theta = \sin^{2} \theta \] Reorganizing this equation leads to: \[ \sin^{2} \theta \cos^{2} \theta + \sin^{4} \theta - \sin^{2} \theta = 0 \] Factoring out \( \sin^{2} \theta \) gives us: \[ \sin^{2} \theta (\cos^{2} \theta + \sin^{2} \theta - 1) = 0 \] Since \( \sin^{2} \theta = 0 \) results in \( \theta = n\pi \) for \( n \in \mathbb{Z} \), we focus on \( \cos^{2} \theta + \sin^{2} \theta - 1 = 0 \), which is actually always true as it represents the identity \( \cos^{2} \theta + \sin^{2} \theta = 1 \). Thus, the complete solutions are \( \theta = n\pi \) where \( n \) is any integer.