How do you verify the identity $\sec x-\cos x=\sin x\tan x$ ?

Hint: We know that cos x and sec x are reciprocal of each other and tan x is the ratio of sin x and cos x. Sum of the square of sin x and cos x is equal to 1. So we can write ${{\sin }^{2}}x=1-{{\cos }^{2}}x$.
These 3 formulas will be enough to prove the identity $\sec x-\cos x=\sin x\tan x$

Complete step by step answer:
We have verify $\sec x-\cos x=\sin x\tan x$
cos x and sec x are reciprocal of each other , so we can write $\sec x=\dfrac{1}{\cos x}$ where cos x is not equal to 0.
We will go from LHS to RHS
$\sec x-\cos x=\dfrac{1}{\cos x}-\cos x$ where cos x is not equal to 0
Further solving we get
$\Rightarrow \sec x-\cos x=\dfrac{1-{{\cos }^{2}}x}{\cos x}$
We know that ${{\sin }^{2}}x=1-{{\cos }^{2}}x$
$\Rightarrow \sec x-\cos x=\dfrac{{{\sin }^{2}}x}{\cos x}$
$\Rightarrow \sec x-\cos x=\sin x\times \dfrac{\sin x}{\cos x}$
We know that $\dfrac{\sin x}{\cos x}=\tan x$
$\Rightarrow \sec x-\cos x=\sin x\tan x$ where cos x is not equal to 0 that implies x is not equal to $\dfrac{n\pi }{2}$ where n is an integer
So we can see that $\sec x-\cos x$ is equal to $\sin x\tan x$

Note:
We also can prove it by multiplying cos x to both LHS and RHS. The LHS will be $1-{{\cos }^{2}}x$ and we know that product of tan x and cos x is equal to sin x, so the RHS will be equal to ${{\cos }^{2}}x$ When writing such identity always mention the domain of x because we can see that $\sec x-\cos x$ is not always equal to $\sin x\tan x$ , sec x and tan x is not defined when x is equal to $\dfrac{n\pi }{2}$ ; n is an integer , so we have to mention it in the answer.