How do you prove $ \dfrac{\cos x}{1+\sin x}=\sec x-\tan x $ ? \[\]

Hint: We begin from the left hand side of the given trigonometric statement and multiply $ \left( 1-\sin x \right) $ in the numerator and denominator. We use Pythagorean trigonometric identity $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 $ in the denominator and separate the numerator to make two new fractions. We simply and use the definition of secant and tangent function in terms of sine and cosine that is $ \sec \theta =\dfrac{1}{\cos \theta },\tan \theta =\dfrac{\sin \theta }{\cos \theta } $ to arrive at the right hand side. \[\]

Complete step-by-step answer:
We know from trigonometry that there are 6 trigonometric function with any angle $ \theta $ as the argument sine $ \left( \sin \theta \right) $ , cosine $ \left( \cos \theta \right) $ , tangent $ \left( \tan \theta \right) $ , cotangent $ \left( \cot \theta \right) $ , secant $ \left( \sec \theta \right) $ and cosecant $ \left( \csc \theta \right) $ . We can convert tangent, cotangent, secant and cosecant trigonometric functions to sine and cosine using the following identities
\[\tan \theta =\dfrac{\sin \theta }{\cos \theta },\cot \theta =\dfrac{\cos \theta }{\sin \theta },\sec \theta =\dfrac{1}{\cos \theta },\csc \theta =\dfrac{1}{\sin \theta }\]
We are given the following statement to prove.
\[\dfrac{\cos x}{1+\sin x}=\sec x-\tan x\]
We see that in the left hand the fractional trigonometric expression is in sine and cosine and the right hand side is in secant and cosecant. So in order to convert to sine, cosine into secant and cosecant we have a denominator in the left hand side from two terms to single term. So we begin from left hand side multiply $ \left( 1-\sin x \right) $ in the numerator and denominator to have
\[\begin{align}
  & \Rightarrow \dfrac{\cos x}{1+\sin x} \\
 & \Rightarrow \dfrac{\cos \left( 1-\sin x \right)}{\left( 1+\sin x \right)\left( 1-\sin x \right)} \\
\end{align}\]
We use the algebraic identity $ \left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}} $ for $ a=1,b=\sin x $ in the denominator of above expression to have;
\[\begin{align}
  & \Rightarrow \dfrac{\cos x\cdot 1-\cos x\cdot \sin x}{{{1}^{2}}-{{\sin }^{2}}x} \\
 & \Rightarrow \dfrac{\cos x-\cos x\cdot \sin x}{1-{{\sin }^{2}}x} \\
\end{align}\]
We use the Pythagorean trigonometric identity $ {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\Rightarrow {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $ for $ \theta =x $ in the above step to have
\[\begin{align}
  & \Rightarrow \dfrac{\cos x-\cos x\sin x}{{{\cos }^{2}}x} \\
 & \Rightarrow \dfrac{\cos x}{{{\cos }^{2}}x}-\dfrac{\cos x\sin x}{{{\cos }^{2}}x} \\
 & \Rightarrow \dfrac{\cos x}{\cos x\cdot \cos x}-\dfrac{\cos x\sin x}{\cos x\cos x} \\
\end{align}\]
We cancel out $ \cos x $ from the numerator and denominator from both the terms in the above expression and the use the definitions $ \sec \theta =\dfrac{1}{\cos \theta },\tan \theta =\dfrac{\sin \theta }{\cos \theta } $ to have
\[\Rightarrow \dfrac{1}{\cos x}-\dfrac{\sin x}{\cos x}=\sec x-\tan x\]
We see that the above expression is at the right hand side of the given statement hence it is proved. \[\]

Note: We note that since a given stamen is well defined $ \sin x+1\ne 0 $ and hence $ x\ne \left( 4n+3 \right)\dfrac{\pi }{2} $ . We also note that $ \sec x,\tan x $ are not defined from $ x=\left( 2n+1 \right)\dfrac{\pi }{2} $ . We can also prove a similar result $ \dfrac{\cos x}{1-\sin x}=\sec x+\tan x $ where we need to multiply $ \left( 1+\sin x \right) $ in the numerator and denominator to convert to secant and tangent.