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