Answer:
The equation is given below as
[tex]\frac{\cos2x}{\cos x}=\cos x-\sin x\tan x[/tex]Step 1:
We will work on the left-hand side, we will have
[tex]\begin{gathered} \cos x-\sin x\tan x \\ \text{recall that,} \\ Quoitent\text{ identity is} \\ \tan x=\frac{\sin x}{\cos x} \end{gathered}[/tex]By substituting the identity above, we will have
[tex]\begin{gathered} \cos x-\sin x\tan x=\cos x-\frac{\sin x.\sin x}{\cos x}=\cos x-\frac{\sin^2x}{\cos x} \\ \end{gathered}[/tex]Here, we will make use of the quotient identity
Step 2:
By writings an expression, we will have
[tex]\begin{gathered} \cos x-\sin x\tan x=\cos x-\frac{\sin x.\sin x}{\cos x} \\ \cos x-\sin x\tan x=\frac{\cos^2x-\sin^2x}{\cos x} \end{gathered}[/tex]Here, we will use the definition of subtraction
[tex]\cos x-\frac{\sin^2x}{\cos x}[/tex]Step 3:
We will apply the double number identity given below
[tex]\begin{gathered} \cos 2\theta=\cos (\theta+\theta)=\cos ^2\theta-\sin ^2\theta \\ \cos 2x=cos(x+x)=\cos ^2x-\sin ^2x \end{gathered}[/tex]By applying this, we will have
[tex]\frac{\cos^2x-\sin^2x}{\cos x}=\frac{\cos2x}{\cos x}[/tex]Here, we will use the double number identity
[tex]\frac{\cos^2x-\sin^2x}{\cos x}[/tex]