Prove the following

$L H S=\sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}$

$=\sqrt{\frac{1}{\cos ^{2} \theta}+\frac{1}{\sin ^{2} \theta}}$ $\left[\because \sec \theta=\frac{1}{\cos \theta}\right.$ and $\left.\operatorname{cosec} \theta=\frac{1}{\sin \theta}\right]$

$=\sqrt{\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin ^{2} \theta \cdot \cos ^{2} \theta}}=\sqrt{\frac{1}{\sin ^{2} \theta \cdot \cos ^{2} \theta}}$ $\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1\right]$

$=\frac{1}{\sin \theta \cdot \cos \theta}=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cdot \cos \theta}$ $\left[\because 1=\sin ^{2} \theta+\cos ^{2} \theta\right]$

$=\frac{\sin ^{2} \theta}{\sin \theta \cdot \cos \theta}+\frac{\cos ^{2} \theta}{\sin \theta \cdot \cos \theta}$

$=\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right.$ and $\left.\cot \theta=\frac{\cos \theta}{\sin \theta}\right]$

$=\tan \theta+\cot \theta=\mathrm{RHS}$