Performance Final Done today. Problem on there I could not figure out.

**mathgeek777** · April 30th 2008, 09:18 AM

I got done with it about 2 hours ago, and there was one problem I could not figure out.

I had to prove that:

$cot x+\frac{sinx}{1+cosx}=cscx$

Could not figure it out to save my life. Could someone show me how to do it?

**Plato** · April 30th 2008, 09:52 AM

$\cot (x) + \frac{{\sin (x)}}{{1 + \cos (x)}} = \frac{{\cos (x)}}{{\sin (x)}} + \frac{{\sin (x)}}{{1 + \cos (x)}}$
$\frac{{\cos (x)}}{{\sin (x)}} + \frac{{\sin (x)}}{{1 + \cos (x)}} = \frac{{\cos (x) + \cos ^2 (x) + \sin ^2 (x)}}{{\sin (x)\left( {1 + \cos (x)} \right)}} = \frac{1}{{\sin (x)}}<br />$

**Soroban** · April 30th 2008, 10:12 AM

Hello, mathgeek777!

Another approach . . .

Prove: . $\cot x+\frac{\sin x}{1+\cos x}\:=\:\csc x$

Multiply the fraction by . $\frac{1-\cos x}{1 - \cos x}$

. . $\cot x \:+ \:\frac{\sin x}{1 + \cos x}\cdot{\color{blue}\frac{1-\cos x}{1-\cos x}} \;\;=\; \;\cot x + \frac{\sin x(1-\cos x)}{1-\cos^2\!x}$

. . $=\;\;\cot x + \frac{\sin x(1-\cos x)}{\sin^2\!x} \;\;=\;\;\cot x + \frac{1-\cos x}{\sin x}$

. . $= \;\;\frac{\cos x}{\sin x} + \frac{1-\cos x}{\sin x} \;\;=\;\;\frac{1}{\sin x} \;\;=\;\;\csc x$

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