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Javaris2011 · Dec 10, 2011, 03:03 PM

sec x csc x/tan x cot x= sin x cos x

corrigan · Dec 10, 2011, 10:54 PM

flat out not true. Counter-example: let x=pi/4.
Then sin x = cos x = 1/√2, sec x = csc x = √2, and tan x =cot x =1.
So sec x csc x/tanx cot x = 2, but sin x cos x = 1/2.

corrigan · Dec 10, 2011, 10:59 PM

didn't come out right.
$x = \frac{\pi}{4}$
$\sin x = \cos x = \frac{1}{\sqrt{2}}$ , $\sec x = \csc x = \sqrt{2}$ , and $\tan x = \cot x =1$ .
So sec x csc x/tanx cot x = 2, but sin x cos x = 1/2.

Aurora2000 · Dec 11, 2011, 08:40 AM

I guess some plus sign is missing, which is common on this site unless "math" environment is used.

I think what Javaris2011 intended was

$\frac{\sec x+\csc x}{\tan x+\cot x}= \frac{\frac{1}{\cos x}+\frac{1}{\sin x}}{\tan x+\frac{1}{\tan x}}= \frac{\cos x+\sin x}{\cos x\sin x}\frac{1}{\tan x+\frac{1}{\tan x}}= \frac{\cos x+\sin x}{\cos x\sin x}\frac{1}{\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}}= \frac{\cos x+\sin x}{\cos x\sin x}\frac{1}{\frac{\sin^2x+\cos^2x}{\cos x\sin x}}= \frac{\cos x+\sin x}{\cos x\sin x} (\cos x\sin x)=\cos x+\sin x$

Aurora2000 · Dec 11, 2011, 08:41 AM

Too long, the last equal sign is followed by $\cos x+\sin x$ , not only $\cos x$