Question
Prove the following.
$(\sin\alpha+\cos\alpha)(\tan\alpha+\cot\alpha)=\sec\alpha+\text{cosec}\alpha$

Answer

LHS $=(\sin\alpha+\cos\alpha)(\tan\alpha+\cot\alpha)$
$=(\sin\alpha+\cos\alpha)\bigg(\frac{\sin\alpha}{\cos\alpha}+\frac{\cos\alpha}{\sin\alpha}\bigg)$ $\bigg[\because\ \tan\theta=\frac{\sin\theta}{\cos\theta}\text{ and }\cot\theta=\frac{\cos\theta}{\sin\theta}\bigg]$
$=(\sin\alpha+\cos\alpha)\bigg(\frac{\sin^2\alpha+\cos^2\alpha}{\sin^\alpha\cdot\cos\alpha}\bigg)$
$=(\sin\alpha+\cos\alpha)\cdot\frac{1}{(\sin\alpha\cdot\cos\alpha)}$ $[\because\ \sin^2\theta+\cos^2\theta=1]$
$=\frac{1}{\cos\alpha}+\frac{1}{\sin\alpha}$ $\Big[\because\ \sec\theta=\frac{1}{\cos\theta}\text{ and cosec}\theta=\frac{1}{\sin\theta}\Big]$
$=\sec\alpha+\cos\alpha=$ RHS

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