Heres the question, a full solution step by step is great as i already know the process to answer a question like this but have trouble doing so.\

$\displaystyle tanx-sinx/tanxsinx = tanxsinx/tanx+sinx$

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- Jun 16th 2008, 01:27 PMFORKTrig
Heres the question, a full solution step by step is great as i already know the process to answer a question like this but have trouble doing so.\

$\displaystyle tanx-sinx/tanxsinx = tanxsinx/tanx+sinx$ - Jun 16th 2008, 01:42 PMMoo
Hi again !

I must guess it's $\displaystyle \frac{\tan x-\sin x}{\tan x \sin x}=\frac{\tan x \sin x}{\tan x+\sin x}$

Please next time, put parenthesis or brackets...

Now, do you have to prove this equation or solve for x ?

And what about showing some of your steps ? (Wink) - Jun 16th 2008, 02:02 PMFORK
- Jun 16th 2008, 03:42 PMtopsquark
$\displaystyle \frac{tan(x) - sin(x)}{tan(x)~sin(x)} = \frac{tan(x)~sin(x)}{tan(x) + sin(x)}$

One way is to start on the left hand side and convert everything to sines and cosines:

$\displaystyle = \frac{\frac{sin(x)}{cos(x)} - sin(x)}{\frac{sin(x)}{cos(x)}~sin(x)}$

Multiply top and bottom by cos(x):

$\displaystyle = \frac{\frac{sin(x)}{cos(x)} - sin(x)}{\frac{sin(x)}{cos(x)}~sin(x)} \cdot \frac{cos(x)}{cos(x)}$

$\displaystyle = \frac{\frac{sin(x)}{cos(x)} \cdot cos(x) - sin(x)~cos(x)}{\frac{sin(x)}{cos(x)}~sin(x) \cdot cos(x)}$

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

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

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

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

Now multiply the top and bottom by tan(x):

$\displaystyle = \frac{sin(x)}{1 + cos(x)} \cdot \frac{tan(x)}{tan(x)}$

$\displaystyle = \frac{tan(x)~sin(x)}{tan(x) + tan(x)~cos(x)}$

$\displaystyle = \frac{tan(x)~sin(x)}{tan(x) + sin(x)}$

-Dan - Jun 16th 2008, 03:50 PMFORK
- Jun 16th 2008, 03:59 PMFORK
- Jun 16th 2008, 04:02 PMtopsquark
$\displaystyle \frac{a}{b} \cdot \frac{c}{c} = \frac{ac}{bc}$ (with suitable restrictions on b and c.) When you multiply fractions you do not need common denominators.

Finishing the problem is easy! C'mon, you need to try a little harder.

$\displaystyle \frac{tan(x)~sin(x)}{tan(x) + tan(x)~cos(x)}$

$\displaystyle = \frac{tan(x)~sin(x)}{tan(x) + \frac{sin(x)}{cos(x)}~cos(x)}$

$\displaystyle = \frac{tan(x)~sin(x)}{tan(x) + sin(x)}$

-Dan