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सिद्ध कीजिए कि $\sqrt{\sec ^{2} \theta+\operatorname{cosec}^{2} \theta}=\tan \theta+\cot \theta$ है।
Solution
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 $\left.\operatorname{cosec} \theta=\frac{1}{\sin \theta}\right]$
$=\sqrt{\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin ^{2} \theta \cdot \cos ^{2} \theta}}=\sqrt{\frac{1}{\sin ^{2} \theta \cdot \cos ^{2} \theta}}$ $\left[\because \cdot \sin ^{2} \theta+\cos ^{2} \theta=1\right]$
$=\frac{1}{\sin \theta \cdot \cos \theta}=\frac{\sin ^{2} \theta+\cos ^{2} \theta}{\sin \theta \cdot \cos \theta}$ $\left[\because 1=\sin ^{2} \theta+\cos ^{2} \theta\right]$
$=\frac{\sin ^{2} \theta}{\sin \theta \cdot \cos \theta}+\frac{\cos ^{2} \theta}{\sin \theta \cdot \cos \theta}$
$=\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}$ $\left[\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right.$ and $\left.\cot \theta=\frac{\cos \theta}{\sin \theta}\right]$
$=\tan \theta+\cot \theta=$ R.H.S.
