Answer :
Alright, let's solve the given equation step by step.
The given equation is:
[tex]\[ \frac{1 + \cos x}{\sin x} + \frac{\sin x}{1 + \cos x} = 4 \][/tex]
First, let's rewrite this equation in a more workable form. Notice that [tex]\(\frac{1 + \cos x}{\sin x}\)[/tex] can be seen as one term and [tex]\(\frac{\sin x}{1 + \cos x}\)[/tex] as another. To find a common denominator, we multiply both sides accordingly.
The next step is to work with a common denominator on the left side:
[tex]\[ \frac{(1 + \cos x)^2 + \sin^2 x}{\sin x (1 + \cos x)} = 4 \][/tex]
Simplify the numerator:
[tex]\[ (1 + \cos x)^2 + \sin^2 x = 1 + 2\cos x + \cos^2 x + \sin^2 x \][/tex]
Using the Pythagorean identity [tex]\(\sin^2 x + \cos^2 x = 1\)[/tex]:
[tex]\[ 1 + 2\cos x + 1 = 2 + 2\cos x \][/tex]
Thus:
[tex]\[ \frac{2 + 2\cos x}{\sin x (1 + \cos x)} = 4 \][/tex]
Now we simplify further:
[tex]\[ \frac{2(1 + \cos x)}{\sin x (1 + \cos x)} = 4 \][/tex]
Here, [tex]\((1 + \cos x)\)[/tex] cancels out from the numerator and the denominator (assuming [tex]\(\cos x \neq -1\)[/tex]):
[tex]\[ \frac{2}{\sin x} = 4 \][/tex]
Thus:
[tex]\[ 2 = 4\sin x \][/tex]
Solving for [tex]\(\sin x\)[/tex]:
[tex]\[ \sin x = \frac{2}{4} = \frac{1}{2} \][/tex]
Therefore, the numerical value of the trigonometric function of [tex]\(x\)[/tex] is [tex]\(\sin x = \frac{1}{2}\)[/tex].
From the given options:
a. [tex]\(\tan x = 2\)[/tex]
b. [tex]\(\sin x = 2\)[/tex]
c. [tex]\(\tan x = \frac{1}{2}\)[/tex]
d. [tex]\(\sin x = \frac{1}{2}\)[/tex]
The correct answer is:
d. [tex]\(\sin x = \frac{1}{2}\)[/tex]
The given equation is:
[tex]\[ \frac{1 + \cos x}{\sin x} + \frac{\sin x}{1 + \cos x} = 4 \][/tex]
First, let's rewrite this equation in a more workable form. Notice that [tex]\(\frac{1 + \cos x}{\sin x}\)[/tex] can be seen as one term and [tex]\(\frac{\sin x}{1 + \cos x}\)[/tex] as another. To find a common denominator, we multiply both sides accordingly.
The next step is to work with a common denominator on the left side:
[tex]\[ \frac{(1 + \cos x)^2 + \sin^2 x}{\sin x (1 + \cos x)} = 4 \][/tex]
Simplify the numerator:
[tex]\[ (1 + \cos x)^2 + \sin^2 x = 1 + 2\cos x + \cos^2 x + \sin^2 x \][/tex]
Using the Pythagorean identity [tex]\(\sin^2 x + \cos^2 x = 1\)[/tex]:
[tex]\[ 1 + 2\cos x + 1 = 2 + 2\cos x \][/tex]
Thus:
[tex]\[ \frac{2 + 2\cos x}{\sin x (1 + \cos x)} = 4 \][/tex]
Now we simplify further:
[tex]\[ \frac{2(1 + \cos x)}{\sin x (1 + \cos x)} = 4 \][/tex]
Here, [tex]\((1 + \cos x)\)[/tex] cancels out from the numerator and the denominator (assuming [tex]\(\cos x \neq -1\)[/tex]):
[tex]\[ \frac{2}{\sin x} = 4 \][/tex]
Thus:
[tex]\[ 2 = 4\sin x \][/tex]
Solving for [tex]\(\sin x\)[/tex]:
[tex]\[ \sin x = \frac{2}{4} = \frac{1}{2} \][/tex]
Therefore, the numerical value of the trigonometric function of [tex]\(x\)[/tex] is [tex]\(\sin x = \frac{1}{2}\)[/tex].
From the given options:
a. [tex]\(\tan x = 2\)[/tex]
b. [tex]\(\sin x = 2\)[/tex]
c. [tex]\(\tan x = \frac{1}{2}\)[/tex]
d. [tex]\(\sin x = \frac{1}{2}\)[/tex]
The correct answer is:
d. [tex]\(\sin x = \frac{1}{2}\)[/tex]