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 February 24th, 2013, 07:08 PM #1 Senior Member   Joined: Aug 2012 From: South Carolina Posts: 866 Thanks: 0 Can't seem to solve... I have spent an hour with the following problem and can get nowhere with it. Somewhere I am missing a big step or something PROBLEM STATES TO VERIFY ALGEBRAICALLY: 5(sec x - tan x)(csc x + 1) = 5 cot x
 February 24th, 2013, 07:37 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Can't seem to solve... My first step would be to rewrite the left side as: $5$$\frac{1-\sin(x)}{\cos(x)}$$$$\frac{1+\sin(x)}{\sin(x)}$$$ Multiply: $5$$\frac{1-\sin^2(x)}{\sin(x)\cos(x)}$$$ Apply Pythagorean identity to numerator: $5$$\frac{\cos^2(x)}{\sin(x)\cos(x)}$$$ Reduce: $5$$\frac{\cos(x)}{\sin(x)}$$=5\cot(x)$
 February 24th, 2013, 07:38 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond Re: Can't seem to solve... \begin{align*}5$$\sec x\,-\,\tan x)(\csc x\,+\,1)\,&=\,5(\sec x\,\csc x\,+\,\sec x\,-\,\sec x\,-\,\tan x) \\ &=\,5(\,\sec x\,\csc x\,-\,\tan x) \\ &=\,5\sec x(\csc x\,-\,\sin x) \\ &=\,5\sec x\(\frac{1\,-\,\sin^2 x}{\sin x}$$ \\ &=\,5\sec x\frac{\cos^2 x}{\sin x} \\ &=\,5\frac{\cos x}{\sin x} \\ &=\,5\cot x\end{align*}
 February 25th, 2013, 07:24 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,942 Thanks: 2210 5(sec x - tan x)(csc x + 1) = 5(sec x - tan x)(sec x + tan x)cot x                                        = 5(sec²x - tan²x)cot x                                        = 5 cot x

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