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

The rule of thumb is to always prove from the side that looks more complicated. I choose the Left-Hand Side (LHS). Look at the bottom of the page for further explanation. $\displaystyle \frac{{\cos x}}{{1-\sin x}}\times \frac{{1+\sin x}}{{1+\sin x}}$

I multiplied both the numerator and denominator by 1-sin x so that I am able to change the denominator to a single expression $\displaystyle {{\cos }^{2}}x$. Use the special algebraic rule $\displaystyle (a+b)(a-b)={{a}^{2}}-{{b}^{2}}$ $\displaystyle \frac{{\cos x(1+\sin x)}}{{1-{{{\sin }}^{2}}x}}$

Use the trigonometric identity $\displaystyle {{\sin }^{2}}x+{{\cos }^{2}}x=1$ $\displaystyle \frac{{\cos x(1+\sin x)}}{{{{{\cos }}^{2}}x}}$ $\displaystyle \frac{{\cos x+\cos x\sin x}}{{{{{\cos }}^{2}}x}}$ $\displaystyle \frac{{\cos x}}{{{{{\cos }}^{2}}x}}+\frac{{\cos x\sin x}}{{{{{\cos }}^{2}}x}}$

Simplify the expression by eliminating $\displaystyle {{\cos }^{2}}x$ from the numerator and denominator. $\displaystyle \frac{1}{{\cos x}}+\frac{{\sin x}}{{\cos x}}$ $\displaystyle \sec x+\tan x$

LHS = RHS

As a rule of thumb, we normally manipulate trigonometric identities from the more ‘complicated’ side. A more ‘complicated’ side has more numbers and variables and it looks more ‘complicated’, it does not mean it is more difficult to manipulate. In fact, there are more ways to manipulate the expression that has more numbers and variables.