# Thread: Help in simplifying equasion

1. ## Help in simplifying equasion

Hi All,
I'm new to this blessed site.
Would appreciate someone's help in simplifying the following equation:

Many thanks !

2. $sec^2(x) = 1+tan^2(x)$,

Can ya finish?

3. $\frac{\tan(x) \sin(x)}{\sec^2(x) - 1} = \frac{\tan(x) \sin(x)}{1 + \tan^2(x) - 1} = \frac{\tan(x) \sin(x)}{\tan^2(x)} = \frac{\sin(x)}{\tan(x)}$.

And $\tan(x) = \frac{\sin(x)}{\cos(x)}$ so we have...

$\frac{\sin(x)}{\tan(x)} = \frac{\sin(x)}{\frac{\sin(x)}{\cos(x)}} = \cos(x)$.

4. If you need clarification on the last step then...

$\frac{\cos(x)}{\cos(x)} = 1$ so...

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

5. Thanks !