## Trigonometry (11th Edition) Clone

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