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Prove the following trigonometric identities:
\( \frac{\sin \theta}{1-\cos \theta}=\operatorname{cosec} \theta+\cot \theta \)
To do:
We have to prove that \( \frac{\sin \theta}{1-\cos \theta}=\operatorname{cosec} \theta+\cot \theta \).
Solution:
We know that,
$\sin ^{2} \theta+cos ^{2} \theta=1$.......(i)
$\operatorname{cosec} \theta=\frac{1}{\sin \theta}$........(ii)
$\cot \theta=\frac{\cos \theta}{\sin \theta}$........(iii)
Therefore,
$\frac{\sin \theta}{1-\cos \theta}=\frac{\sin \theta}{1-\cos \theta}\times \frac{1+\cos \theta}{1+\cos \theta}$ (Multiply and divide by $1+\cos \theta$)
$=\frac{(\sin \theta)(1+\cos \theta)}{(1-\cos \theta)(1+\cos \theta)}$
$=\frac{\sin \theta(1+\cos \theta)}{1^2-\cos^2 \theta)}$
$=\frac{\sin \theta(1+\cos \theta)}{\sin^2 \theta}$ (From (i))
$=\frac{1+\cos \theta}{\sin \theta}$
$=\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}$
$=\operatorname{cosec} \theta+\cot \theta$ (From (ii) and (iii))
Hence proved.