Question
Prove the following identities:
$\sin\theta(1+\tan\theta)+\cos\theta(1+\cot\theta)=(\sec\theta+\text{cosec }\theta)$
$\sin\theta(1+\tan\theta)+\cos\theta(1+\cot\theta)=(\sec\theta+\text{cosec }\theta)$
✓
Answer
$\text{L.H.S.}=\sin\theta(1+\tan\theta)+\cos\theta(1+\cot\theta)$
$=\Big(\sin\theta+\frac{\sin^2\theta}{\cos\theta}\Big)+\cos\theta\Big(1+\frac{\cos\theta}{\sin\theta}\Big)$
$=\Big(\sin\theta+\frac{\sin^2\theta}{\cos\theta}\Big)+\Big(\cos\theta+\frac{\cos^2\theta}{\sin\theta}\Big)$
$=\Big(\frac{\sin\theta\cos\theta+\sin^2\theta}{\cos\theta}\Big)+\Big(\frac{\cos\theta\sin\theta+\cos^2\theta}{\sin\theta}\Big)$
$=\frac{\sin\theta\cos\theta}{\cos\theta}+\frac{\sin^2\theta}{\cos\theta}+\frac{\cos\theta\sin\theta}{\sin\theta}+\frac{\cos^2\theta}{\sin\theta}$
$=\frac{\sin\theta\cos\theta}{\cos\theta}+\frac{\cos\theta\sin\theta}{\sin\theta}+\frac{\sin^2\theta}{\cos\theta}+\frac{\cos^2\theta}{\sin\theta}$
$=\sin\theta\cos\theta\Big(\frac{\sin\theta+\cos\theta}{\cos\theta\sin\theta}\Big)+\frac{\sin^3\theta+\cos^3\theta}{\cos\theta\sin\theta}$
$=\sin\theta\cos\theta\Big(\frac{\sin\theta+\cos\theta}{\cos\theta\sin\theta}\Big)\\+\frac{(\sin\theta+\cos\theta)\big(\sin^2\theta-\sin\theta\cos\theta+\cos^2\theta\big)}{\cos\theta\sin\theta}$ $\Big[\because\text{a}^3+\text{b}^3=(\text{a}+\text{b})\big(\text{a}^2-\text{ab}+\text{b}^2\big)\Big]$
$=(\sin\theta+\cos\theta)\Big[\frac{\sin\theta\cos\theta}{\cos\theta\sin\theta}+\frac{(1-\sin\theta\cos\theta)}{\cos\theta\sin\theta}\Big]$
$=(\sin\theta+\cos\theta)\Big[\frac{\sin\theta\cos\theta+1-\sin\theta\cos\theta}{\cos\theta\sin\theta}\Big]$
$=(\sin\theta+\cos\theta)\Big(\frac{1}{\cos\theta\sin\theta}\Big)$
$=\frac{\sin}{\cos\theta\sin\theta}+\frac{\cos\theta}{\cos\theta\sin\theta}=\sec\theta+\text{cosec }\theta$
$=\text{R.H.S.}$
$\therefore\ \text{L.H.S.}=\text{R.H.S}$
$=\Big(\sin\theta+\frac{\sin^2\theta}{\cos\theta}\Big)+\cos\theta\Big(1+\frac{\cos\theta}{\sin\theta}\Big)$
$=\Big(\sin\theta+\frac{\sin^2\theta}{\cos\theta}\Big)+\Big(\cos\theta+\frac{\cos^2\theta}{\sin\theta}\Big)$
$=\Big(\frac{\sin\theta\cos\theta+\sin^2\theta}{\cos\theta}\Big)+\Big(\frac{\cos\theta\sin\theta+\cos^2\theta}{\sin\theta}\Big)$
$=\frac{\sin\theta\cos\theta}{\cos\theta}+\frac{\sin^2\theta}{\cos\theta}+\frac{\cos\theta\sin\theta}{\sin\theta}+\frac{\cos^2\theta}{\sin\theta}$
$=\frac{\sin\theta\cos\theta}{\cos\theta}+\frac{\cos\theta\sin\theta}{\sin\theta}+\frac{\sin^2\theta}{\cos\theta}+\frac{\cos^2\theta}{\sin\theta}$
$=\sin\theta\cos\theta\Big(\frac{\sin\theta+\cos\theta}{\cos\theta\sin\theta}\Big)+\frac{\sin^3\theta+\cos^3\theta}{\cos\theta\sin\theta}$
$=\sin\theta\cos\theta\Big(\frac{\sin\theta+\cos\theta}{\cos\theta\sin\theta}\Big)\\+\frac{(\sin\theta+\cos\theta)\big(\sin^2\theta-\sin\theta\cos\theta+\cos^2\theta\big)}{\cos\theta\sin\theta}$ $\Big[\because\text{a}^3+\text{b}^3=(\text{a}+\text{b})\big(\text{a}^2-\text{ab}+\text{b}^2\big)\Big]$
$=(\sin\theta+\cos\theta)\Big[\frac{\sin\theta\cos\theta}{\cos\theta\sin\theta}+\frac{(1-\sin\theta\cos\theta)}{\cos\theta\sin\theta}\Big]$
$=(\sin\theta+\cos\theta)\Big[\frac{\sin\theta\cos\theta+1-\sin\theta\cos\theta}{\cos\theta\sin\theta}\Big]$
$=(\sin\theta+\cos\theta)\Big(\frac{1}{\cos\theta\sin\theta}\Big)$
$=\frac{\sin}{\cos\theta\sin\theta}+\frac{\cos\theta}{\cos\theta\sin\theta}=\sec\theta+\text{cosec }\theta$
$=\text{R.H.S.}$
$\therefore\ \text{L.H.S.}=\text{R.H.S}$
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