Prove that,$\tan \theta + \tan (90^{\circ} - \theta) = \sec \theta \sec (90^{\circ} - \theta)$

(N/A) $L$.$H$.$S$. $= \tan \theta + \tan (90^{\circ} - \theta)$
Since $\tan (90^{\circ} - \theta) = \cot \theta$,we have:
$= \tan \theta + \cot \theta$
$= \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}$
$= \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta}$
Since $\sin^2 \theta + \cos^2 \theta = 1$,we get:
$= \frac{1}{\sin \theta \cos \theta}$
$= \frac{1}{\cos \theta} \cdot \frac{1}{\sin \theta}$
$= \sec \theta \operatorname{cosec} \theta$
Since $\operatorname{cosec} \theta = \sec (90^{\circ} - \theta)$,we get:
$= \sec \theta \sec (90^{\circ} - \theta) = \text{R.H.S.}$