Question
Prove the following trigonometric identities.
$\frac{\sec\theta+1}{\sec\theta+1}=\Big(\frac{\sin\theta}{1+\cos\theta}\Big)^2$
$\frac{\sec\theta+1}{\sec\theta+1}=\Big(\frac{\sin\theta}{1+\cos\theta}\Big)^2$
✓
Answer
We have
$\text{L.H.S}=\frac{\sec\theta+1}{\sec\theta+1}=\frac{\frac{1}{\cos\theta}-1}{\frac{1}{\cos\theta}+1}$
$=\frac{\frac{1-\cos\theta}{\cos\theta}}{\frac{1+\cos\theta}{\cos\theta}}$
$=\frac{1-\cos\theta}{1+\cos\theta}$
Multiplying both the numerator and the denominator by $(1+\cos\theta),$ we get
$\frac{\sec\theta-1}{\sec\theta+1}=\frac{(1-\cos\theta)(1+\cos\theta)}{(1+\cos\theta)(1+\cos\theta)}$
$=\frac{(1-\cos^2\theta)}{(1+\cos\theta)^2}$
$=\frac{\sin^2\theta}{(1+\cos\theta)^2}$
$=\Big(\frac{\sin\theta}{1+\cos\theta}\Big)^2=\text{R.H.S}$
$\therefore \text{L.H.S}=\text{R.H.S}$
$\text{L.H.S}=\frac{\sec\theta+1}{\sec\theta+1}=\frac{\frac{1}{\cos\theta}-1}{\frac{1}{\cos\theta}+1}$
$=\frac{\frac{1-\cos\theta}{\cos\theta}}{\frac{1+\cos\theta}{\cos\theta}}$
$=\frac{1-\cos\theta}{1+\cos\theta}$
Multiplying both the numerator and the denominator by $(1+\cos\theta),$ we get
$\frac{\sec\theta-1}{\sec\theta+1}=\frac{(1-\cos\theta)(1+\cos\theta)}{(1+\cos\theta)(1+\cos\theta)}$
$=\frac{(1-\cos^2\theta)}{(1+\cos\theta)^2}$
$=\frac{\sin^2\theta}{(1+\cos\theta)^2}$
$=\Big(\frac{\sin\theta}{1+\cos\theta}\Big)^2=\text{R.H.S}$
$\therefore \text{L.H.S}=\text{R.H.S}$
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