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