Let's solve the given mathematical expression step by step:
The problem states:
[tex]\[
\frac{\cos x}{1 - \sin x} + \frac{\cos x}{1 + \sin x} = \csc x
\][/tex]
To begin with, let's simplify the left-hand side (LHS) of the equation:
[tex]\[
\frac{\cos x}{1 - \sin x} + \frac{\cos x}{1 + \sin x}
\][/tex]
To combine these fractions, we need a common denominator. The common denominator of [tex]\(1 - \sin x\)[/tex] and [tex]\(1 + \sin x\)[/tex] is:
[tex]\[
(1 - \sin x)(1 + \sin x)
\][/tex]
First, recall the algebraic identity:
[tex]\[
(1 - \sin x)(1 + \sin x) = 1 - \sin^2 x
\][/tex]
Additionally, using the Pythagorean identity, we know:
[tex]\[
1 - \sin^2 x = \cos^2 x
\][/tex]
Now, we can express each fraction with the common denominator:
[tex]\[
\frac{\cos x (1 + \sin x)}{\cos^2 x} + \frac{\cos x (1 - \sin x)}{\cos^2 x}
\][/tex]
Expanding the numerators, we have:
[tex]\[
\frac{\cos x + \cos x \sin x}{\cos^2 x} + \frac{\cos x - \cos x \sin x}{\cos^2 x}
\][/tex]
Combining the terms in the numerator, we get:
[tex]\[
\frac{\cos x + \cos x \sin x + \cos x - \cos x \sin x}{\cos^2 x}
\][/tex]
Note that [tex]\( \cos x \sin x \)[/tex] and [tex]\(- \cos x \sin x\)[/tex] cancel out:
[tex]\[
\frac{2 \cos x}{\cos^2 x}
\][/tex]
We can simplify this fraction:
[tex]\[
\frac{2 \cos x}{\cos^2 x} = \frac{2}{\cos x}
\][/tex]
So the simplified form of the LHS is:
[tex]\[
\frac{2}{\cos x}
\][/tex]
Next, we compare this to the right-hand side (RHS) of the original equation:
[tex]\[
\csc x = \frac{1}{\sin x}
\][/tex]
Now, let's consider the trigonometric identity of [tex]\(\csc x = \frac{1}{\sin x}\)[/tex] and see if:
[tex]\[
\frac{2}{\cos x} = \frac{1}{\sin x}
\][/tex]
For these two expressions to be equal, it must be true that:
[tex]\[
\frac{2}{\cos x} = \frac{1}{\sin x}
\][/tex]
Cross-multiplying provides the relationship:
[tex]\[
2 \sin x = \cos x
\][/tex]
This equation does not hold true for all values of [tex]\(x\)[/tex], thus the original equation is not generally true:
The left-hand side simplifies to:
[tex]\[
\frac{2}{\cos x}
\][/tex]
And the right-hand side is:
[tex]\[
\frac{1}{\sin x}
\][/tex]
Since these two are not equal in general, the equality:
[tex]\[
\frac{\cos x}{1 - \sin x} + \frac{\cos x}{1 + \sin x} = \csc x
\][/tex]
is false.
