Answer:
[tex]\sin x(1+\tan ^2x)[/tex][tex]\sin x\sec ^2x[/tex][tex]\sin x\cdot\frac{1}{\cos^2x}[/tex][tex]\frac{\sin x}{\cos x}\cdot\frac{1}{\cos x}[/tex]
Explanation:
Given the below;
[tex]\sin x+\sin x\tan ^2x=\tan x\sec x[/tex]
We'll go ahead and manipulate the left-hand side of the equation following the below steps;
Step 1: Factor out sinx;
[tex]\sin x(1+\tan ^2x)=\tan x\sec x[/tex]
Step 2:
Recall the below trig identity;
[tex]\sec ^2x=1+\tan ^2x[/tex]
Substituting the above, we'll have;
[tex]\sin x\sec ^2x=\tan x\sec x[/tex]
Step 3:
Recall that;
[tex]\begin{gathered} \sec x=\frac{1}{\cos x} \\ \sec ^2x=\frac{1}{\cos ^2x} \end{gathered}[/tex]
So we'll have;
[tex]\sin x\cdot\frac{1}{\cos^2x}=\tan x\sec x[/tex]
Step 4:
We can further simplify as seen below;
[tex]\begin{gathered} \sin x\cdot\frac{1}{\cos x\cdot\cos x}=\tan x\sec x \\ \frac{\sin x}{\cos x}\cdot\frac{1}{\cos x}=\tan x\sec x \end{gathered}[/tex]
Step 5:
Recall that;
[tex]\begin{gathered} \tan x=\frac{\sin x}{\cos x} \\ \sec x=\frac{1}{\cos x} \end{gathered}[/tex]
So we can rewrite our equation as;
[tex]\tan x\sec x=\tan x\sec x[/tex]
