trig proof help

**Tweety** · November 4th 2009, 03:51 AM

Show that;

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

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

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

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

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

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

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

Thank you!

**mathaddict** · November 4th 2009, 05:22 AM

Originally Posted by Tweety

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

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

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

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

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

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

Thank you!

HI

*** I will continue from here .

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

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

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

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

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

$ =2\tan x $

**Tweety** · November 4th 2009, 07:46 AM

Originally Posted by mathaddict

HI

*** I will continue from here .

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

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

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

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

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

$ =2\tan x $

hey thanks for that but I am not sure how you got 1-sinx as the denomninator?

Could you explain ,

thanks

**Beard** · November 4th 2009, 08:08 AM

$\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

**Tweety** · November 4th 2009, 10:26 AM

Originally Posted by Beard

$\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?

**Grandad** · November 4th 2009, 12:28 PM

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

$\frac{\left(\dfrac{A}{B}\right)}{\left(\dfrac{C}{B }\right)}$

you can multiply top-and-bottom of the main fraction by $B$ :

$\frac{\left(\dfrac{A}{B}\right)\times B}{\left(\dfrac{C}{B}\right)\times B}$

which then cancels each of the $'B'$ denominators to leave you with

$\frac{A}{C}$

Grandad

**Tweety** · November 4th 2009, 01:02 PM

Originally Posted by Beard

$\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

Originally Posted by Grandad

Hello TweetyYes. When you have something like

$\frac{\left(\dfrac{A}{B}\right)}{\left(\dfrac{C}{B }\right)}$

you can multiply top-and-bottom of the main fraction by $B$ :

$\frac{\left(\dfrac{A}{B}\right)\times B}{\left(\dfrac{C}{B}\right)\times B}$

which then cancels each of the $'B'$ denominators to leave you with

$\frac{A}{C}$

Grandad

Thank you.

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