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