সমাধান কর : $\sec ^{2} \theta+\tan ^{2} \theta=\frac{5}{3}, 0<\theta<\pi$

$-\frac{\pi}{6},-\frac{5 \pi}{6}$

$-\frac{\pi}{6}, \frac{5 \pi}{6}$

$\frac{\pi}{6},-\frac{5 \pi}{6}$

$\frac{\pi}{6}, \frac{5 \pi}{6}$