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.\
$\displaystyle tanx-sinx/tanxsinx = tanxsinx/tanx+sinx$
$\displaystyle \frac{tan(x) - sin(x)}{tan(x)~sin(x)} = \frac{tan(x)~sin(x)}{tan(x) + sin(x)}$
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
$\displaystyle \frac{a}{b} \cdot \frac{c}{c} = \frac{ac}{bc}$ (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