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# Same Question, Different Mistake

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Are there any other mistakes that you would expect students to make on this problem?

Thanks to Kristen Fouss for the submission.

## 4 replies on “Same Question, Different Mistake”

1 + tan^2 = sec^2, so taking the square root of both sides, 1 + tan = sec.

$dfrac {tan left( xright) } {1+dfrac {1} {ccos left( xright) }}+dfrac {1+dfrac {1} {ccos left( xright) }}{tan left( xright) }$
$dfrac {sin left( xright) } {1}+dfrac {1} {sin left( xright) }$
and possibly $dfrac {sin ^{2}left( xright) +1} {sin(x)}$
and why not $dfrac {cos^{2}left( xright) } {sin left( xright) }=dfrac {cosleft( xright) } {tan left( xright) }$

(kids, this is NOT CORRECT)
My Friday latex workout/ Phew!

Wah!
$dfrac {tan left( xright) } {1+dfrac {1} {ccos left( xright) }}+dfrac {1+dfrac {1} {ccos left( xright) }}{tan left( xright) }$
$dfrac {sin left( xright) } {1}+dfrac {1} {sin left( xright) }$
and possibly $dfrac {sin ^{2}left( xright) +1} {sin(x)}$
and why not $dfrac {cos^{2}left( xright) } {sin left( xright) }=dfrac {cosleft( xright) } {tan left( xright) }$