Prove that

$\mathrm{LHS}=\frac{\cot ^{2} \theta(\sec \theta-1)}{(1+\sin \theta)}+\frac{\sec ^{2} \theta(\sin \theta-1)}{(1+\sec \theta)}$

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

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

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

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

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

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

$=0$

$=\mathrm{RHS}$