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