Question
Prove the following trigonometric identities.
$(\sin\theta+\cos\theta)(\tan\theta+\cos\theta)=\sec\theta+\text{cosec }\theta$
$(\sin\theta+\cos\theta)(\tan\theta+\cos\theta)=\sec\theta+\text{cosec }\theta$
✓
Answer
Consider the L.H.S $(\sin\theta+\cos\theta)(\tan\theta+\cot\theta)$$\text{L.H.S}=(\sin\theta+\cos\theta)(\tan\theta+\cot\theta)$
$=(\sin\theta+\cos\theta)\Big(\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}\Big)$ $=(\sin\theta+\cos\theta)\Big(\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\times\cos\theta}\Big)$ $=\frac{\sin\theta+\cos\theta}{\sin\theta\times\cos\theta} \big[\sin^2\theta+\cos^2\theta=1\Big]$ $=\frac{1}{\cos\theta}+\frac{1}{\sin\theta}$ $=\sec\theta+\text{cosec }\theta=\text{R.H.S}$ $\therefore\ \text{L.H.S}=\text{R.H.S}$
$=(\sin\theta+\cos\theta)\Big(\frac{\sin\theta}{\cos\theta}+\frac{\cos\theta}{\sin\theta}\Big)$ $=(\sin\theta+\cos\theta)\Big(\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\times\cos\theta}\Big)$ $=\frac{\sin\theta+\cos\theta}{\sin\theta\times\cos\theta} \big[\sin^2\theta+\cos^2\theta=1\Big]$ $=\frac{1}{\cos\theta}+\frac{1}{\sin\theta}$ $=\sec\theta+\text{cosec }\theta=\text{R.H.S}$ $\therefore\ \text{L.H.S}=\text{R.H.S}$
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