$\displaystyle \frac{\sec{x}\tan{x}}{\tan^3{x}} = \frac{\cos{x}}{\sin{x}} $ the part in the middle is what I am missing
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Hello, TYTY! $\displaystyle \frac{\sec x\tan x}{\tan^3\!x} \:=\: \frac{\cos x}{\sin x}$ the part in the middle is what I am missing . . . . oh, is that all? Please check the problem again . . . The given statement is not an identity.
Originally Posted by TYTY $\displaystyle \frac{\sec{x}\tan{x}}{\tan^3{x}} = \frac{\cos{x}}{\sin{x}} $ the part in the middle is what I am missing Hi TYTY, Maybe, you meant $\displaystyle \frac{\sec{x}\tan{x}}{\tan^3{x}} = \frac{\cos{x}}{\sin^2{x}}$
Last edited by masters; Mar 25th 2009 at 07:59 AM.
ouch better start that whole problem over again...
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