#### Answer

$$\frac{\sin t}{1+\cos t}=\frac{1-\cos t}{\sin t}$$
As in the Work Step by Step, prove that 2 sides are equal and it follows that the equation is an identity.

#### Work Step by Step

$$\frac{\sin t}{1+\cos t}=\frac{1-\cos t}{\sin t}$$
You can choose any side to start with in this exercise. I choose the left one to start with here.
$$X=\frac{\sin t}{1+\cos t}$$
The trick here is to multiply both numerator and denominator by $1-\cos t$.
$$X=\frac{\sin t(1-\cos t)}{(1+\cos t)(1-\cos t)}$$
$$X=\frac{\sin t(1-\cos t)}{1-\cos^2t}$$ (since $(A-B)(A+B)=A^2-B^2$)
- Now, according to Pythagorean identity: $\sin^2t=1-\cos^2t$
$$X=\frac{\sin t(1-\cos t)}{\sin^2t}$$
$$X=\frac{1-\cos t}{\sin t}$$
Therefore, $$\frac{\sin t}{1+\cos t}=\frac{1-\cos t}{\sin t}$$
So the equation is an identity.