Prove that: $\frac{sin\theta}{cot\theta+cosec\theta}=2+\frac{sin\theta}{cot\theta-cosec\theta}$.

Given: $\frac{sin\theta}{cot\theta+cosec\theta}=2+\frac{sin\theta}{cot\theta-cosec\theta}$.

To do: To prove $L.H.S.=R.H.S.$

$L.H.S. =\frac{sin\theta}{cot\theta+cosec\theta}$

$= \frac{sin\theta}{\frac{cos\theta}{sin\theta}+\frac{1}{sin\theta}}$

$= \frac{sin\theta}{\frac{cos\theta+1}{sin\theta}}$

$= \frac{sin^{2}\theta}{(1+cos\theta)}$

$= \frac{(1-cos^{2}\theta)}{(1+cos\theta)}$                            $\because sin^{2}\theta = 1 - cos^{2}\theta$

$= \frac{[( 1+cos\theta)( 1-cos\theta)]}{( 1+cos\theta)}$                $\because a^{2}-b^{2} = ( a+b)( a-b)$

$= 1 - cos\theta ......( 1)$

$R.H.S. = 2+ \frac{sin\theta}{(cot\theta-cosec\theta)}$

$= 2+\frac{sin\theta}{\frac{cos\theta}{sin\theta}-\frac{1}{sin\theta}}$

$= 2+\frac{sin\theta}{\frac{cos\theta-1}{sin\theta}}$

$= 2+ \frac{[ sin^{2}\theta}{(cos\theta-1)]}$

$= 2 - \frac{( sin^{2}\theta)}{( 1-cos\theta)}$

$= 2- \frac{[ ( 1-cos^{2}\theta)}{( 1-cos\theta)]}$

$= 2-\frac{[( 1+cos\theta)( 1-cos\theta)}{( 1-cos\theta)]}$

$= 2 - ( 1+cos\theta)$

$= 2-1-cos\theta$

$= 1-cos\theta ......(2)$

Form $( 1)$ & $( 2)$ , we conclude that,

$( 1) = ( 2)$

$L.H.S. = R.H.S.$