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