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hines57 · Dec 12, 2011, 11:26 PM

1/secx-tanx=secx tanx

LuckyChucky13 · Dec 12, 2011, 11:29 PM

Is the 1 divided by the secx alone or is it over secx - tanx together?

Aurora2000 · Dec 13, 2011, 03:10 AM

$\frac{1}{\sec x-\tan x}=\sec x+\tan x$

Multiply both sides by $\sec x-\tan x$ and this reads

$1= \sec^2 x-\tan^2 x$

which is solved by direct computation

$\sec^2 x-\tan^2 x = \frac{1}{\cos^2x}-\frac{\sin^2x}{\cos^2x}=\frac{1-\sin^2x}{\cos^2x}=1$

hines57 · Dec 13, 2011, 09:39 AM

The is over secx-tanx together because the is the common denominator