How do you simplify $ \dfrac{\cos x}{\sec x+\tan x} $ ?

Hint: We will use the relation between trigonometric functions for simplifying the given expression. We will write the terms in the denominator in terms of the sine function and cosine function. Then we will use the identity $ {{\cos }^{2}}x+{{\sin }^{2}}x=1 $ to rewrite the numerator. After that we will factorize the numerator using the identity $ {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) $ and obtain the simplification of the given expression.

Complete step by step answer:
We have to simplify the expression $ f\left( x \right)=\dfrac{\cos x}{\sec x+\tan x} $ . Now, using the relations between the trigonometric functions, we know that $ \tan x=\dfrac{\sin x}{\cos x} $ and $ \sec x=\dfrac{1}{\cos x} $ . Substituting these values in the given expression, we get the following,
 $ \begin{align}
  & f\left( x \right)=\dfrac{\cos x}{\dfrac{1}{\cos x}+\dfrac{\sin x}{\cos x}} \\
 & \therefore f\left( x \right)=\dfrac{\cos x}{\dfrac{1+\sin x}{\cos x}} \\
\end{align} $
We know that $ \dfrac{\left( \dfrac{a}{b} \right)}{\left( \dfrac{c}{d} \right)}=\dfrac{a}{b}\times \dfrac{d}{c} $ . Using this, we can write the above expression as,
 $ \begin{align}
  & f\left( x \right)=\cos x\times \dfrac{\cos x}{1+\sin x} \\
 & \therefore f\left( x \right)=\dfrac{{{\cos }^{2}}x}{1+\sin x} \\
\end{align} $
We have the following identity, $ {{\cos }^{2}}x+{{\sin }^{2}}x=1 $ . Therefore, we can write $ {{\cos }^{2}}x=1-{{\sin }^{2}}x $ . Substituting this value in the numerator, we get the following,
 $ f\left( x \right)=\dfrac{1-{{\sin }^{2}}x}{1+\sin x} $
We have the algebraic identity, $ {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) $ . Using this identity, we can factorize the numerator as $ 1-{{\sin }^{2}}x=\left( 1+\sin x \right)\left( 1-\sin x \right) $ . Substituting this value in place of the numerator, we get,
 $ \begin{align}
  & f\left( x \right)=\dfrac{\left( 1+\sin x \right)\left( 1-\sin x \right)}{1+\sin x} \\
 & \therefore f\left( x \right)=1-\sin x \\
\end{align} $
Therefore, we obtain the simplification of the given expression as $ 1-\sin x $ .

Note:
It is important to know the relations between trigonometric functions for solving this type of question. We should also know the trigonometric identities as they are useful in the simplification of trigonometric expressions. It is essential that we write each step explicitly while simplifying the given expression. In this way, we can notice the terms or parts of the expression that can be substituted using identities or relations. Also, we can avoid making mistakes and obtain the correct simplification needed.