How do you prove $\sec x-\cos x=\sin x\tan x$ ?

Hint: We are given that right side is sin x tan x and left side is sec x – cos x, we will first learn how we can simplify sec x, tan x, into the simplest form then we use $\sec x=\dfrac{1}{\cos x},\tan x=\dfrac{\sin x}{\cos x}$ . We will also need the identity ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ . We will simplify both the right side and the left side one by one and see that they are the same .

Complete step by step answer:
We are asked to prove that $\sec x-\cos x$ is the same as $\sin x\tan x$ .
To prove it, we will learn how ratios are ratios connected to each other.
In our problem, we have 4 ratios so that are $\sin x,\cos x,\sec x,\tan x$ .
So, we learn how they are connected. We know that $\sin x=\dfrac{1}{\cos ecx}$ , $\cos x=\dfrac{1}{\sec x}$ or say $\sec x=\dfrac{1}{\cos x}$ , $\tan x$ is given as $\dfrac{\sin x}{\cos x}$ and lastly ${{\sin }^{2}}x+{{\cos }^{2}}x=1$ , we can change this to get ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ or ${{\cos }^{2}}x=1-{{\sin }^{2}}x$ as needed.
Now we will verify that $\sec x-\cos x=\sin x\tan x$ .
So, we consider the left hand side.
We have $\sec x-\cos x$ .
As we know that $\sec x=\dfrac{1}{\cos x}$
So, $\sec x-\cos x=\dfrac{1}{\cos x}-\cos x$ .By simplifying, we get –
$=\dfrac{1}{\cos x}-\dfrac{\cos x}{1}$ .
By LCM we simplify further and we get –
$=\dfrac{1-{{\cos }^{2}}x}{\cos x}$ .
We know that ${{\cos }^{2}}x+{{\sin }^{2}}x=1$
So, $1-{{\cos }^{2}}x={{\sin }^{2}}x$ .
So using the above identity, we get –
$\sec x-\cos x=\dfrac{{{\sin }^{2}}x}{\cos x}$ …………………………… (1)
So we get left side simplified as $\dfrac{{{\sin }^{2}}x}{\cos x}$
Now we consider the right side, we have $\sin x\times \tan x$ .
As $\tan x=\dfrac{\sin x}{\cos x}$ , so
$\sin x\tan x=\sin x\times \dfrac{\sin x}{\cos x}$
Simplifying, we get –
$=\dfrac{{{\sin }^{2}}x}{\cos x}$
So, $\sin x\tan x=\dfrac{{{\sin }^{2}}x}{\cos x}$ …………………………… (2)
From eq (1) and (2) we get –
$\sin x\tan x$ and $\sec x-\cos x$ are equal to one another.

Hence $\sec x-\cos x=\sin x\tan x$

Note: We can also extend the proof of the left hand side to reach to the right hand side but many time it get complicated so we normally just simplify to the simplest term possible then start working on the other side we left our left side at $\sec x-\cos x=\dfrac{{{\sin }^{2}}x}{\cos x}$ as ${{\sin }^{2}}x=\sin x\sin x$ .
So, using this, we get –
$=\dfrac{\sin x\sin x}{\cos x}$
$=\sin x\times \left( \dfrac{\sin x}{\cos x} \right)$
As $\dfrac{\sin x}{\cos x}=\tan x$
So, $=\sin x\times \tan x$
= Right Hand Side.
Hence proved.