Answer:
Option d. [tex]sin(x)=\frac{1}{2}[/tex]
Step-by-step explanation:
The complete question is
Given (1+cosx)/(sinx) + (sinx)/(1+cosx) =4, find a numerical value of one trigonometric function of x.
a. tanx=2
b. sinx=2
c. tanx=1/2
d. sinx=1/2
we have
[tex]\frac{1+cos(x)}{sin(x)}+\frac{sin(x)}{1+cos(x)}=4[/tex]
Find the common denominator and adds the fractions
[tex]\frac{(1+cos(x))^2+sin^2(x)}{(1+cos(x))sin(x)}=4[/tex]
Expanded the numerator
[tex]\frac{(1+2cos(x)+cos^2(x)+sin^2(x)}{(1+cos(x))sin(x)}=4[/tex]
Remember that
[tex]sin^2(x)+cos^2(x)=1[/tex] ----> trigonometric identity
substitute
[tex]\frac{1+2cos(x)+1}{(1+cos(x))sin(x)}=4[/tex]
[tex]\frac{2+2cos(x)}{(1+cos(x))sin(x)}=4[/tex]
Factor 2 in the numerator
[tex]\frac{2(1+cos(x))}{(1+cos(x))sin(x)}=4[/tex]
Simplify
[tex]\frac{2}{sin(x)}=4[/tex]
[tex]sin(x)=\frac{1}{2}[/tex]
