How would you prove or disprove #cotx - cosx/cotx = cos^2x/(1 + sinx)#?

Question

How would you prove or disprove #cotx - cosx/cotx = cos^2x/(1  + sinx)#?

Anna Andrews · Accepted Answer

Rewrite all terms in #cotx# as #cosx/sinx#.

#cosx/sinx - cosx/(cosx/sinx) = cos^2x/(1 + sinx)#

#cosx/sinx - cosx xx sinx/cosx = cos^2x/(1 + sinx)#

#cosx/sinx - sinx = cos^2x/(1 + sinx)#

Rewrite the right-hand side using the identity #sin^2x + cos^2x = 1#.

#cosx/sinx - sinx = (1 - sin^2x)/(1 + sinx)#

#cosx/sinx - sinx = ((1 + sinx)(1 - sinx))/(1 + sinx)#

#cosx/sinx - sinx = 1 - sinx#

#(cosx - sin^2x)/sinx = 1 - sinx#

The identity is false, because no matter what you do with the left hand side, you will never be able to get on the right.

Hopefully this helps!

Josiah Anderson · Answer

undefined To prove: $\frac{\cot x - \cos x}{\cot x} = \frac{\cos^2 x}{1 + \sin x}$

First, express both sides in terms of sine and cosine.

Left side: 
$\frac{\cot x - \cos x}{\cot x} = \frac{\frac{\cos x}{\sin x} - \cos x}{\frac{\cos x}{\sin x}} = \frac{\cos x - \cos^2 x}{\cos x} = 1 - \cos x$

Right side: 
$\frac{\cos^2 x}{1 + \sin x} = \frac{\cos^2 x}{1 + \sin x} \times \frac{1 - \sin x}{1 - \sin x} = \frac{\cos^2 x (1 - \sin x)}{1 - \sin^2 x} = \frac{\cos^2 x (1 - \sin x)}{\cos^2 x} = 1 - \sin x$

Since both sides simplify to $1 - \cos x$ and $1 - \sin x$, the expression is proven.