To show that [tex]\(\frac{\sin x \tan x}{1 - \cos x} = 1 + \sec x\)[/tex], we will perform a detailed step-by-step algebraic manipulation.
First, let's rewrite the left-hand side (LHS):
[tex]\[
\frac{\sin x \tan x}{1 - \cos x}
\][/tex]
Recall that the tangent function [tex]\(\tan x\)[/tex] can be written in terms of sine and cosine:
[tex]\[
\tan x = \frac{\sin x}{\cos x}
\][/tex]
Substituting [tex]\(\tan x\)[/tex] in the expression, we get:
[tex]\[
\frac{\sin x \left(\frac{\sin x}{\cos x}\right)}{1 - \cos x} = \frac{\sin^2 x}{\cos x (1 - \cos x)}
\][/tex]
Next, let's simplify the right-hand side (RHS):
[tex]\[
1 + \sec x
\][/tex]
Recall that the secant function [tex]\(\sec x\)[/tex] is the reciprocal of the cosine function:
[tex]\[
\sec x = \frac{1}{\cos x}
\][/tex]
So, the RHS becomes:
[tex]\[
1 + \frac{1}{\cos x}
\][/tex]
To combine these terms into a single fraction, we get a common denominator:
[tex]\[
1 + \frac{1}{\cos x} = \frac{\cos x}{\cos x} + \frac{1}{\cos x} = \frac{\cos x + 1}{\cos x}
\][/tex]
Now, we need to show that both expressions (LHS and RHS) are equal:
[tex]\[
\frac{\sin^2 x}{\cos x (1 - \cos x)} = \frac{\cos x + 1}{\cos x}
\][/tex]
To proceed, let's manipulate the LHS ([tex]\(\frac{\sin^2 x}{\cos x (1 - \cos x)}\)[/tex]) further. Recognize the Pythagorean identity:
[tex]\[
\sin^2 x = 1 - \cos^2 x
\][/tex]
Thus, replacing [tex]\(\sin^2 x\)[/tex] with [tex]\(1 - \cos^2 x\)[/tex], we get:
[tex]\[
\frac{1 - \cos^2 x}{\cos x (1 - \cos x)}
\][/tex]
Next, factor the numerator [tex]\(1 - \cos^2 x\)[/tex]:
[tex]\[
1 - \cos^2 x = (1 - \cos x)(1 + \cos x)
\][/tex]
So, we have:
[tex]\[
\frac{(1 - \cos x)(1 + \cos x)}{\cos x (1 - \cos x)}
\][/tex]
We observe that the [tex]\((1 - \cos x)\)[/tex] terms cancel out:
[tex]\[
\frac{1 + \cos x}{\cos x}
\][/tex]
This matches precisely with our simplified RHS:
[tex]\[
\frac{\cos x + 1}{\cos x}
\][/tex]
Therefore:
[tex]\[
\frac{\sin x \tan x}{1 - \cos x} = 1 + \sec x
\][/tex]
Thus, we have shown that:
[tex]\[
\frac{\sin x \tan x}{1 - \cos x} = 1 + \sec x
\][/tex]
as required.
