Sure! Let's solve the equation step-by-step:
[tex]\[
\tan x + \frac{\cos x}{1 + \sin x} = \frac{1}{\cos x}
\][/tex]
First, let's rewrite the trigonometric identities involved for a better understanding.
[tex]\[
\tan x = \frac{\sin x}{\cos x}
\][/tex]
So, the given equation can be rewritten as:
[tex]\[
\frac{\sin x}{\cos x} + \frac{\cos x}{1 + \sin x} = \frac{1}{\cos x}
\][/tex]
Now, let's multiply every term by [tex]\(\cos x\)[/tex] to clear the denominators:
[tex]\[
\sin x + \frac{\cos^2 x}{1 + \sin x} = 1
\][/tex]
Next, let's isolate the fraction on one side:
[tex]\[
\frac{\cos^2 x}{1 + \sin x} = 1 - \sin x
\][/tex]
Now, multiply both sides by [tex]\((1 + \sin x)\)[/tex] to eliminate the fraction:
[tex]\[
\cos^2 x = (1 - \sin x)(1 + \sin x)
\][/tex]
Notice that [tex]\((1 - \sin x)(1 + \sin x)\)[/tex] is a difference of squares:
[tex]\[
\cos^2 x = 1 - \sin^2 x
\][/tex]
Recall the Pythagorean identity:
[tex]\[
\cos^2 x + \sin^2 x = 1
\][/tex]
Thus:
[tex]\[
\cos^2 x = \cos^2 x
\][/tex]
This tells us that the equation holds true without offering new information, indicating there might be an issue with the constraints of the given equation and its solutions.
To investigate this, let's check if there are any specific solutions. Trying values or analyzing the derived results, we get the indication that there are no specific [tex]\(x\)[/tex] values within the constraints of the functions under these trigonometric relationships to hold the equation true in a novel sense.
Therefore, we conclude that:
[tex]\[
\boxed{\text{No solutions}}
\][/tex]
