Prove that $\sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}=\tan \theta+\cot \theta$

(A) $L.H.S. = \sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}$
$= \sqrt{\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta}}$ $\left[\because \sec \theta = \frac{1}{\cos \theta} \text{ and } \operatorname{cosec} \theta = \frac{1}{\sin \theta}\right]$
$= \sqrt{\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin ^{2} \theta \cdot \cos ^{2} \theta}} = \sqrt{\frac{1}{\sin ^{2} \theta \cdot \cos ^{2} \theta}}$ $\left[\because \sin ^{2} \theta+\cos ^{2} \theta = 1\right]$
$= \frac{1}{\sin \theta \cdot \cos \theta} = \frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cdot \cos \theta}$ $\left[\because 1 = \sin ^{2} \theta+\cos ^{2} \theta\right]$
$= \frac{\sin ^{2} \theta}{\sin \theta \cdot \cos \theta} + \frac{\cos ^{2} \theta}{\sin \theta \cdot \cos \theta}$
$= \frac{\sin \theta}{\cos \theta} + \frac{\cos \theta}{\sin \theta}$ $\left[\because \tan \theta = \frac{\sin \theta}{\cos \theta} \text{ and } \cot \theta = \frac{\cos \theta}{\sin \theta}\right]$
$= \tan \theta + \cot \theta = R.H.S.$