Question

Given 1+cosx\sinx+sinx\1+cosx=4, find a numerical value of one trigonometric function of x.

Answer 1

Answer:

Option d. $sin(x)=\frac{1}{2}$

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

$\frac{1+cos(x)}{sin(x)}+\frac{sin(x)}{1+cos(x)}=4$

Find the common denominator and adds the fractions

$\frac{(1+cos(x))^2+sin^2(x)}{(1+cos(x))sin(x)}=4$

Expanded the numerator

$\frac{(1+2cos(x)+cos^2(x)+sin^2(x)}{(1+cos(x))sin(x)}=4$

Remember that

$sin^2(x)+cos^2(x)=1$ ----> trigonometric identity

substitute

$\frac{1+2cos(x)+1}{(1+cos(x))sin(x)}=4$

$\frac{2+2cos(x)}{(1+cos(x))sin(x)}=4$

Factor 2 in the numerator

$\frac{2(1+cos(x))}{(1+cos(x))sin(x)}=4$

Simplify

$\frac{2}{sin(x)}=4$

$sin(x)=\frac{1}{2}$