I got done with it about 2 hours ago, and there was one problem I could not figure out.
I had to prove that:
$\displaystyle cot x+\frac{sinx}{1+cosx}=cscx$
Could not figure it out to save my life. Could someone show me how to do it?
I got done with it about 2 hours ago, and there was one problem I could not figure out.
I had to prove that:
$\displaystyle cot x+\frac{sinx}{1+cosx}=cscx$
Could not figure it out to save my life. Could someone show me how to do it?
$\displaystyle \cot (x) + \frac{{\sin (x)}}{{1 + \cos (x)}} = \frac{{\cos (x)}}{{\sin (x)}} + \frac{{\sin (x)}}{{1 + \cos (x)}}$
$\displaystyle \frac{{\cos (x)}}{{\sin (x)}} + \frac{{\sin (x)}}{{1 + \cos (x)}} = \frac{{\cos (x) + \cos ^2 (x) + \sin ^2 (x)}}{{\sin (x)\left( {1 + \cos (x)} \right)}} = \frac{1}{{\sin (x)}}
$
Hello, mathgeek777!
Another approach . . .
Prove: .$\displaystyle \cot x+\frac{\sin x}{1+\cos x}\:=\:\csc x$
Multiply the fraction by .$\displaystyle \frac{1-\cos x}{1 - \cos x}$
. . $\displaystyle \cot x \:+ \:\frac{\sin x}{1 + \cos x}\cdot{\color{blue}\frac{1-\cos x}{1-\cos x}} \;\;=\; \;\cot x + \frac{\sin x(1-\cos x)}{1-\cos^2\!x}$
. . $\displaystyle =\;\;\cot x + \frac{\sin x(1-\cos x)}{\sin^2\!x} \;\;=\;\;\cot x + \frac{1-\cos x}{\sin x}$
. . $\displaystyle = \;\;\frac{\cos x}{\sin x} + \frac{1-\cos x}{\sin x} \;\;=\;\;\frac{1}{\sin x} \;\;=\;\;\csc x$