Question:
$\frac{\cos \theta \operatorname{cosec} \theta-\sin \theta \sec \theta}{\cos \theta+\sin \theta}=\operatorname{cosec} \theta-\sec \theta$
Solution:
$\mathrm{LHS}=\frac{\cos \theta \operatorname{cosec} \theta-\sin \theta \sec \theta}{\cos \theta+\sin \theta}$
$=\frac{\frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta}}{\cos \theta+\sin \theta}$
$=\frac{\cos ^{2} \theta-\sin ^{2} \theta}{\cos \theta \sin \theta(\cos \theta+\sin \theta)}$
$=\frac{(\cos \theta+\sin \theta)(\cos \theta-\sin \theta)}{\cos \theta \sin \theta(\cos \theta+\sin \theta)}$
$=\frac{(\cos \theta-\sin \theta)}{\cos \theta \sin \theta}$
$=\frac{1}{\sin \theta}-\frac{1}{\cos \theta}$
$=\operatorname{cosec} \theta-\sec \theta$
$=$ RHS
Hence, LHS $=\mathrm{RHS}$