0) Let \( g(x)=\frac{\sin (x)}{1+\cos (x)}+\frac{1+\cos (x)}{\sin (x)} \) We would prefer one fraction. Show that \( \frac{2}{\sin (x)} \) is an equivalent formula for \( g(x) \).

To show that \( g(x) = \frac{2}{\sin(x)} \), we start with the expression for \( g(x) \): \[ g(x) = \frac{\sin(x)}{1 + \cos(x)} + \frac{1 + \cos(x)}{\sin(x)} \] To combine these two fractions into a single fraction, we need a common denominator. The common denominator will be \( (1 + \cos(x)) \sin(x) \). We rewrite each term with this common denominator: 1. The first term becomes: \[ \frac{\sin(x)}{1 + \cos(x)} = \frac{\sin^2(x)}{(1 + \cos(x)) \sin(x)} \] 2. The second term becomes: \[ \frac{1 + \cos(x)}{\sin(x)} = \frac{(1 + \cos(x))^2}{(1 + \cos(x)) \sin(x)} \] Now we can combine the two fractions: \[ g(x) = \frac{\sin^2(x) + (1 + \cos(x))^2}{(1 + \cos(x)) \sin(x)} \] Next, we simplify the numerator: \[ (1 + \cos(x))^2 = 1 + 2\cos(x) + \cos^2(x) \] Thus, the numerator becomes: \[ \sin^2(x) + 1 + 2\cos(x) + \cos^2(x) \] Using the Pythagorean identity \( \sin^2(x) + \cos^2(x) = 1 \), we can simplify: \[ \sin^2(x) + \cos^2(x) = 1 \] So the numerator simplifies to: \[ 1 + 1 + 2\cos(x) = 2 + 2\cos(x) \] Now we can rewrite \( g(x) \): \[ g(x) = \frac{2 + 2\cos(x)}{(1 + \cos(x)) \sin(x)} \] Factoring out a 2 from the numerator gives: \[ g(x) = \frac{2(1 + \cos(x))}{(1 + \cos(x)) \sin(x)} \] Assuming \( 1 + \cos(x) \neq 0 \) (which is true except for specific values of \( x \)), we can cancel \( 1 + \cos(x) \) from the numerator and denominator: \[ g(x) = \frac{2}{\sin(x)} \] Thus, we have shown that: \[ g(x) = \frac{2}{\sin(x)} \] This completes the proof that \( g(x) \) is equivalent to \( \frac{2}{\sin(x)} \).