Question:
Prove the following trigonometric identities.
$\frac{1-\cos \theta}{\sin \theta}=\frac{\sin \theta}{1+\cos \theta}$
Solution:
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 numerator and denominator by $(1+\cos \theta)$, we have
$\frac{1-\cos \theta}{\sin \theta}=\frac{(1-\cos \theta)(1+\cos \theta)}{\sin \theta(1+\cos \theta)}$
$=\frac{1-\cos ^{2} \theta}{\sin \theta(1+\cos \theta)}$
$=\frac{\sin ^{2} \theta}{\sin \theta(1+\cos \theta)}$
$=\frac{\sin \theta}{1+\cos \theta}$
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