Certainly! Let's solve the equation step-by-step to determine if the left-hand side is equivalent to the right-hand side.
Given the equation:
[tex]\[
(\sin x + \tan x) \left( \frac{\sin x}{1 + \cos x} \right) = \sin x \cdot \tan x
\][/tex]
First, let's simplify the left-hand side (LHS) of the equation. We start by distributing the term [tex]\(\frac{\sin x}{1 + \cos x}\)[/tex]:
[tex]\[
(\sin x + \tan x) \left( \frac{\sin x}{1 + \cos x} \right) = \sin x \left( \frac{\sin x}{1 + \cos x} \right) + \tan x \left( \frac{\sin x}{1 + \cos x} \right)
\][/tex]
We know that [tex]\(\tan x\)[/tex] can be written in terms of [tex]\(\sin x\)[/tex] and [tex]\(\cos x\)[/tex]:
[tex]\[
\tan x = \frac{\sin x}{\cos x}
\][/tex]
Substitute [tex]\(\tan x\)[/tex] into the equation:
[tex]\[
\sin x \left( \frac{\sin x}{1 + \cos x} \right) + \left( \frac{\sin x}{\cos x} \right) \left( \frac{\sin x}{1 + \cos x} \right)
\][/tex]
Simplify the fractions:
[tex]\[
\frac{\sin^2 x}{1 + \cos x} + \frac{\sin^2 x}{\cos x (1 + \cos x)}
\][/tex]
Combine the terms over a common denominator:
[tex]\[
= \frac{\sin^2 x \cos x + \sin^2 x}{\cos x (1 + \cos x)}
= \frac{\sin^2 x (\cos x + 1)}{\cos x (1 + \cos x)}
\][/tex]
Notice that [tex]\((\cos x + 1)\)[/tex] appears in both the numerator and denominator, so they cancel each other out:
[tex]\[
= \frac{\sin^2 x}{\cos x}
\][/tex]
Thus, our left-hand side simplifies to:
[tex]\[
\frac{\sin^2 x}{\cos x}
\][/tex]
Now, let's simplify the right-hand side (RHS) of the equation:
[tex]\[
\sin x \cdot \tan x = \sin x \cdot \frac{\sin x}{\cos x} = \frac{\sin^2 x}{\cos x}
\][/tex]
We have shown that both the left-hand side and the right-hand side of the equation simplify to:
[tex]\[
\frac{\sin^2 x}{\cos x}
\][/tex]
Since both sides are equal, we can conclude that the original equation is true. Thus:
[tex]\[
(\sin x + \tan x) \left( \frac{\sin x}{1 + \cos x} \right) = \sin x \cdot \tan x
\][/tex]
is a true statement.
