(sinx/1+cosx) + (sinx/1-cosx). i need help starting this to solve?):

Question

jim_thompson5910 · Answer

$$\Large \frac{\sin(x)}{1+\cos(x)} + \frac{\sin(x)}{1-\cos(x)}$$

$$\Large \frac{\sin(x)(1-\cos(x))}{(1+\cos(x))(1-\cos(x))} + \frac{\sin(x)}{1-\cos(x)}$$


$$\Large \frac{\sin(x)(1-\cos(x))}{(1+\cos(x))(1-\cos(x))} + \frac{\sin(x)(1+\cos(x))}{(1+\cos(x))(1-\cos(x))}$$

does that help?

anonymous · Answer

yes:) then i got $$\frac{\sin^2x(1-cosx)(1+cosx) }{(1-cosx)(1+cosx) } $$ would it just end up being $$\sin^2x$$

jim_thompson5910 · Answer

close, but you would do this

$$\Large \frac{\sin(x)}{1+\cos(x)} + \frac{\sin(x)}{1-\cos(x)}$$ 

$$\Large \frac{\sin(x)(1-\cos(x))}{(1+\cos(x))(1-\cos(x))} + \frac{\sin(x)}{1-\cos(x)}$$ 

$$\Large \frac{\sin(x)(1-\cos(x))}{(1+\cos(x))(1-\cos(x))} + \frac{\sin(x)(1+\cos(x))}{(1+\cos(x))(1-\cos(x))}$$

$$\Large \frac{\sin(x)(1-\cos(x))+\sin(x)(1+\cos(x))}{(1+\cos(x))(1-\cos(x))}$$

$$\Large \frac{\sin(x)-\sin(x)\cos(x)+\sin(x)+\sin(x)\cos(x)}{(1+\cos(x))(1-\cos(x))}$$

$$\Large \frac{2\sin(x)}{(1+\cos(x))(1-\cos(x))}$$

$$\Large \frac{2\sin(x)}{1-\cos^2(x)}$$

$$\Large \frac{2\sin(x)}{\sin^2(x)}$$

$$\Large \frac{2}{\sin(x)}$$

$$\Large 2\csc(x)$$

anonymous · Answer

Oh! I had just cancelled the top and bottom two factors. Thank you so much! it really helped me!:)

jim_thompson5910 · Answer

glad it did