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