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DominiRican1013
 one year ago
The expression (secx + tanx)2 is the same as _____.
DominiRican1013
 one year ago
The expression (secx + tanx)2 is the same as _____.

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(\sec x + \tan x)^2\] Remember how to expand out a squared expression like this? :)

DominiRican1013
 one year ago
Best ResponseYou've already chosen the best response.0\[(\frac{ 1 }{ cosx }+\frac{ sinx }{ cosx})^2 ?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Before we continue, is there by chance choices?

DominiRican1013
 one year ago
Best ResponseYou've already chosen the best response.0yes hold on...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you're able to get past that first part, here is a hint to help you finish it up. Remember your Pythagorean identity for tangent, \[\sec ^2x=1+\tan ^2x\]

DominiRican1013
 one year ago
Best ResponseYou've already chosen the best response.0\[A. 1+2\tan^2+2secx tanx\] \[B. 1+2cscx\] \[C. \sec^2x+\tan^2x\] \[D. \sec^2x+2cscx+\tan^2x\]

DominiRican1013
 one year ago
Best ResponseYou've already chosen the best response.0These are the choices

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Well after you expand you would get \[\sec^2(x)+2\sec(x)\tan(x)+\tan^2(x)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But remember after you expand everything out, you can replace the sec^2x(that you'll end up with) with 1+tan^2x.

DominiRican1013
 one year ago
Best ResponseYou've already chosen the best response.0I'm so lost What I have so far is 1+tan(x)+2sec(x)tan(x)+1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Expand the original equation like you would a binomial

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0except in this case instead of (a+b)^2 A is Sec (x) and B is Tan (x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0look at my previous posts, they are the exact steps on how to do this.
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