Prove that $(\sin^{4} \theta - \cos^{4} \theta + 1) \operatorname{cosec}^{2} \theta = 2$.

(N/A) $L$.$H$.$S$. $= (\sin^{4} \theta - \cos^{4} \theta + 1) \operatorname{cosec}^{2} \theta$
$= [(\sin^{2} \theta - \cos^{2} \theta)(\sin^{2} \theta + \cos^{2} \theta) + 1] \operatorname{cosec}^{2} \theta$
Since $\sin^{2} \theta + \cos^{2} \theta = 1$,we have:
$= (\sin^{2} \theta - \cos^{2} \theta + 1) \operatorname{cosec}^{2} \theta$
Substitute $1 - \cos^{2} \theta = \sin^{2} \theta$:
$= (\sin^{2} \theta + \sin^{2} \theta) \operatorname{cosec}^{2} \theta$
$= (2 \sin^{2} \theta) \operatorname{cosec}^{2} \theta$
$= 2 (\sin^{2} \theta \cdot \operatorname{cosec}^{2} \theta)$
Since $\sin \theta \cdot \operatorname{cosec} \theta = 1$,we get:
$= 2(1) = 2 = \text{R.H.S.}$