# simplify

• Dec 28th 2011, 09:48 AM
Duke
simplify
I want to go from $ac/(a^4sin^4(x)-c^2a^2sin^2(x))^{0.5}$ to $cosec^2(x)/((a/c)^2-1-cot^2(x))^{0.5}$
• Dec 28th 2011, 10:26 AM
Amer
Re: simplify
Quote:

Originally Posted by Duke
I want to go from $ac/(a^4sin^4(x)-c^2a^2sin^2(x))^{0.5}$ to $cosec^2(x)/((a/c)^2-1-cot^2(x))^{0.5}$

$\frac{ac}{\sqrt{a^4 \sin ^4 (x) - c^2 a^2 \sin ^2 (x)}}$
to
$\frac{csc ^2 (x)}{\sqrt{\frac{a^2}{c^2} -1 - \cot ^2 (x)}}$

you can divide the denominator and nominator by ac note $ac = \sqrt{a^2c^2}$
take $\sin ^4 (x)$ at the square root denominator as a factor
• Dec 28th 2011, 12:23 PM
Duke
Re: simplify
I take sin^4(x) out and that helps numerator but I can't see the way forward.
• Dec 28th 2011, 12:59 PM
Siron
Re: simplify
We are given:
$\frac{ac}{\sqrt{a^4\sin^4(x)-c^2a^2\sin^2(x)}}$

Rewriting as:
$\frac{ac}{\sqrt{a^2c^2\sin^4(x)\left[\frac{a^2}{c^2}-\frac{1}{\sin^2(x)}\right]}}$
$=\frac{ac}{ac\sin^2(x)\sqrt{\left(\frac{a}{c} \right)^2-\frac{\sin^2(x)+\cos^2(x)}{\sin^2(x)}}}$
$=\frac{1}{\sin^2(x)\sqrt{\left(\frac{a}{c}\right)^ 2-\frac{\sin^2(x)}{\sin^2(x)}-\frac{\cos^2(x)}{\sin^2(x)}}}$
$=\frac{\csc^2(x)}{\sqrt{\left(\frac{a}{c}\right)^2-1-\cot^2(x)}}$