Help in simplifying equasion

**naorca** · November 5th 2009, 01:51 PM

Hi All,
I'm new to this blessed site.

Would appreciate someone's help in simplifying the following equation:

Many thanks !

**Deadstar** · November 5th 2009, 01:57 PM

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

Can ya finish?

**naorca** · November 5th 2009, 02:12 PM

**Deadstar** · November 5th 2009, 02:31 PM

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

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

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

**Deadstar** · November 5th 2009, 02:37 PM

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

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

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

**naorca** · November 5th 2009, 03:02 PM

Thanks !

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