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