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#1

$Old$ November 5th, 2009, 02:51 PM

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$Default$ 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 !

Last edited by naorca; November 6th, 2009 at 03:36 AM.

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#2

$Old$ November 5th, 2009, 02:57 PM

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

Can ya finish?

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$Old$ November 5th, 2009, 03:12 PM

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$Old$ November 5th, 2009, 03:31 PM

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

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

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

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$Old$ November 5th, 2009, 03:37 PM

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

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

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

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$Old$ November 5th, 2009, 04:02 PM

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Thanks !

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