1. ## trig proof help

Show that;

$\displaystyle \frac{cotx}{cosecx-1} -\frac{cosx}{1+sinx} = 2tanx$
$\displaystyle \frac {\frac{cosx}{sinx}}{ \frac{1}{sinx}-1} - \frac{cosx}{1+sinx}$

$\displaystyle \frac{ \frac{cosx}{sinx}}{ \frac{1-sinx}{sinx} }- \frac{cosx}{1+sinx}$

$\displaystyle \frac{cosx}{sinx} \times \frac{sinx}{1-sinx} = \frac{cosx sinx}{sinx-sin^2x}$

Than can I cancel out the sinx's? so I get;

$\displaystyle \frac{cosx}{-sinx} - \frac{cosx}{1+sinx} = 2tanx$

Than I am not sure where to go from here, any suggestions appreciated.

Thank you!

2. Originally Posted by Tweety
$\displaystyle \frac {\frac{cosx}{sinx}}{ \frac{1}{sinx}-1} - \frac{cosx}{1+sinx}$

$\displaystyle \frac{ \frac{cosx}{sinx}}{ \frac{1-sinx}{sinx} }- \frac{cosx}{1+sinx}$ ***

$\displaystyle \frac{cosx}{sinx} \times \frac{sinx}{1-sinx} = \frac{cosx sinx}{sinx-sin^2x}$

Than can I cancel out the sinx's? so I get;

$\displaystyle \frac{cosx}{-sinx} - \frac{cosx}{1+sinx} = 2tanx$

Than I am not sure where to go from here, any suggestions appreciated.

Thank you!
HI

*** I will continue from here .

$\displaystyle \frac{\cos x}{1-\sin x}-\frac{\cos x}{1+\sin x}$

$\displaystyle =\frac{\cos x(1+\sin x-\cos x(1-\sin x))}{1-\sin^2 x}$

$\displaystyle =\frac{\cos x=\cos x\sin x-\cos x+\cos x\sin x}{\cos^2 x}$

$\displaystyle =\frac{2\cos x\sin x}{\cos^2 x}$

$\displaystyle =2\frac{\sin x}{\cos x}$

$\displaystyle =2\tan x$

HI

*** I will continue from here .

$\displaystyle \frac{\cos x}{1-\sin x}-\frac{\cos x}{1+\sin x}$

$\displaystyle =\frac{\cos x(1+\sin x-\cos x(1-\sin x))}{1-\sin^2 x}$

$\displaystyle =\frac{\cos x=\cos x\sin x-\cos x+\cos x\sin x}{\cos^2 x}$

$\displaystyle =\frac{2\cos x\sin x}{\cos^2 x}$

$\displaystyle =2\frac{\sin x}{\cos x}$

$\displaystyle =2\tan x$
hey thanks for that but I am not sure how you got 1-sinx as the denomninator?

Could you explain ,

thanks

4. $\displaystyle \frac{ \frac{cosx}{sinx}}{ \frac{1-sinx}{sinx} }- \frac{cosx}{1+sinx}$

When at this point, the sinx as denominators (in the first fraction) cancel each other out leaving the 1-sin(x) as the overall denominator

5. Originally Posted by Beard
$\displaystyle \frac{ \frac{cosx}{sinx}}{ \frac{1-sinx}{sinx} }- \frac{cosx}{1+sinx}$

When at this point, the sinx as denominators (in the first fraction) cancel each other out leaving the 1-sin(x) as the overall denominator
Thank you, Also is that a general rule that when you have one fraction on top of another fraction and thier denominators are the same you can cancel them?

6. Hello Tweety
Originally Posted by Tweety
Thank you, Also is that a general rule that when you have one fraction on top of another fraction and thier denominators are the same you can cancel them?
Yes. When you have something like
$\displaystyle \frac{\left(\dfrac{A}{B}\right)}{\left(\dfrac{C}{B }\right)}$
you can multiply top-and-bottom of the main fraction by $\displaystyle B$:
$\displaystyle \frac{\left(\dfrac{A}{B}\right)\times B}{\left(\dfrac{C}{B}\right)\times B}$
which then cancels each of the $\displaystyle 'B'$ denominators to leave you with
$\displaystyle \frac{A}{C}$

7. Originally Posted by Beard
$\displaystyle \frac{ \frac{cosx}{sinx}}{ \frac{1-sinx}{sinx} }- \frac{cosx}{1+sinx}$

When at this point, the sinx as denominators (in the first fraction) cancel each other out leaving the 1-sin(x) as the overall denominator
$\displaystyle \frac{\left(\dfrac{A}{B}\right)}{\left(\dfrac{C}{B }\right)}$
you can multiply top-and-bottom of the main fraction by $\displaystyle B$:
$\displaystyle \frac{\left(\dfrac{A}{B}\right)\times B}{\left(\dfrac{C}{B}\right)\times B}$
which then cancels each of the $\displaystyle 'B'$ denominators to leave you with
$\displaystyle \frac{A}{C}$