Question
Prove that $\sqrt{\sec^2\theta+\text{cosec}^2\theta}=\tan\theta+\cot\theta.$
✓
Answer
LHS $=\sqrt{\sec^2\theta+\text{cosec}^2\theta}$
$=\sqrt{\frac{1}{\cos^2\theta}+\frac{1}{\sin^2\theta}}=\sqrt{\frac{\sin^2\theta+\cos^2\theta}{\cos^2\theta\sin^2\theta}}$
$=\sqrt{\frac{1}{\sin^2\theta\cdot\cos^2\theta}}=\frac{1}{\sin\theta\cos\theta}$
$=\text{cosec}\theta\sec\theta$
RHS $=\tan\theta+\cot\theta $
$=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}$
$=\frac{1}{\sin\theta\cdot\cos\theta}$
$=\text{cosec}\theta\cdot\sec\theta=$ LHS
Hence,proved.
$=\sqrt{\frac{1}{\cos^2\theta}+\frac{1}{\sin^2\theta}}=\sqrt{\frac{\sin^2\theta+\cos^2\theta}{\cos^2\theta\sin^2\theta}}$
$=\sqrt{\frac{1}{\sin^2\theta\cdot\cos^2\theta}}=\frac{1}{\sin\theta\cos\theta}$
$=\text{cosec}\theta\sec\theta$
RHS $=\tan\theta+\cot\theta $
$=\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}$
$=\frac{1}{\sin\theta\cdot\cos\theta}$
$=\text{cosec}\theta\cdot\sec\theta=$ LHS
Hence,proved.
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