$=\frac{1}{\sin x}\bullet \frac{1}{\cos x}$ (Algebra simplification)

$=\csc x\sec x$ (Reciprocal identities)

$\frac{{\csc }^{2}x}{\cot x}$

$\csc \theta -\sin \theta =\cot \theta \cos \theta$

$=\frac{1}{\sin \theta }-\sin \theta$ (Reciprocal identity)

$=\frac{1}{\sin \theta }-\frac{\sin \theta }{1}\bullet \frac{\sin \theta }{\sin \theta }$ (Common denominator)

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

$=\frac{{(\cos \theta )}^2}{\sin \theta }$ (Pythagorean identity)

$=\frac{\cos \theta }{\sin \theta }\bullet \frac{\cos \theta }{1}$ (Algebra)

= $\cot \theta \cos \theta$ (Ratio identity)

$\csc \theta -\sin \theta$

$\sec x=\frac{\cot x\tan x}{\cos x}$

$=\frac{\frac{\cos x}{\sin x}\bullet \frac{\sin x}{\cos x}}{\cos x}$

$=\frac{1}{\cos x}$

$=\sec x$

$\frac{\cot x\tan x}{\cos x}$

$\frac{\cos x}{1-\sin x}=\sec x+\tan x$

$=\frac{\cos x}{1-\sin x}\bullet \frac{1+\sin x}{1+\sin x}$

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

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

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

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

$=\sec x+\tan x$

$\frac{\cos x}{1-\sin x}$

$\tan x+\cot x=\csc x\sec x$

$=\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}$

$=\frac{{\sin }^{2}x+{\cos }^{2}x}{\cos x\sin x}$

$=\frac{1}{\cos x\sin x}$

$=\frac{1}{\cos x}\bullet \frac{1}{\sin x}$

$=\sec x\csc x$

$\tan x+\cot x$