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hines57 · Dec 12, 2011, 09:58 PM

((secx-tanx)^2 1)/(secxcscx-tanxcscx)=2tanx please help me with this if you can its got me stumped

Aurora2000 · Dec 13, 2011, 03:28 AM

$((\sec x-\tan x)^2+1)/(\sec x\csc x-\tan x \csc x) = 2\tan x$

By direct computation:

Numerator:
$(\sec x-\tan x)^2+1 =\sec^2 x+\tan^2 x-2\sec x\tan x+1 =<br /> \frac{1}{\cos^2 x}+\frac{\sin^2x}{\cos^2x}-2\frac{1}{\cos x}\frac{\sin x}{\cos x}+1=$
$\frac{1+\sin^2 x}{\cos^2 x}- \frac{2\sin x}{\cos^2 x}+1= \frac{1+\sin ^2 x-2\sin x}{\cos^2 x}+1=$
$\frac{(1-\sin x)^2}{\cos^2 x} +1 = \frac{2(1-\sin x)}{\cos^2 x}$

Denominator:

$\sec x\csc x-\tan x \csc x = \frac{1}{\sin x\cos x}- \frac{\sin x}{\cos x}\frac{1}{\sin x}<br /> =\frac{1}{\sin x\cos x} -\frac{1}{\cos x} =\frac{1-\sin x}{\sin x\cos x}$

Then the original fraction reads

$\frac{ \frac{2(1-\sin x)}{\cos^2 x} }{\frac{1-\sin x}{\sin x\cos x}}<br /> = \frac{2(1-\sin x)}{\cos^2 x} \frac{\sin x\cos x}{1-\sin x}=$
$2\frac{\sin x}{\cos x}=2\tan x$