Question
Prove the following trigonometric identities.
$\frac{\cos^2\theta}{\sin\theta}-\text{cosec}\theta+\sin\theta=0$
$\frac{\cos^2\theta}{\sin\theta}-\text{cosec}\theta+\sin\theta=0$
✓
Answer
We have to prove $\frac{\cos^2\theta}{\sin\theta}-\text{cosec}\theta+\sin\theta=0$
We know that, $\sin^2\theta+\cos^2\theta=1$
So,
$\text{L.H.S}=\frac{\cos^2\theta}{\sin\theta}-\text{cosec}\theta+\sin\theta$
$=\Big(\frac{\cos^2\theta}{\sin\theta}-\text{cosec}\theta\Big)+\sin\theta$
$=\Big(\frac{\cos^2\theta}{\sin\theta}-\frac{1}{\sin\theta}\Big)+\sin\theta$
$=\Big(\frac{\cos^2\theta-1}{\sin\theta}\Big)+\sin\theta$
$=\Big(\frac{-\sin^2\theta}{\sin\theta}\Big)+\sin\theta$
$=-\sin\theta+\sin\theta$
$=0=\text{R.H.S}$
$\therefore\ \text{L.H.S.}=\text{R.H.S}$
We know that, $\sin^2\theta+\cos^2\theta=1$
So,
$\text{L.H.S}=\frac{\cos^2\theta}{\sin\theta}-\text{cosec}\theta+\sin\theta$
$=\Big(\frac{\cos^2\theta}{\sin\theta}-\text{cosec}\theta\Big)+\sin\theta$
$=\Big(\frac{\cos^2\theta}{\sin\theta}-\frac{1}{\sin\theta}\Big)+\sin\theta$
$=\Big(\frac{\cos^2\theta-1}{\sin\theta}\Big)+\sin\theta$
$=\Big(\frac{-\sin^2\theta}{\sin\theta}\Big)+\sin\theta$
$=-\sin\theta+\sin\theta$
$=0=\text{R.H.S}$
$\therefore\ \text{L.H.S.}=\text{R.H.S}$
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