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I simplified $1 + \sec^2 x$ to

$1 + 1/\cos^2 x$ then

$2/\cos^2 x$ but I am stuck here.

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    $\begingroup$ $\sin^2 x + \cos^2 x=1 $ , so : $C$ $\endgroup$
    – Pedja
    Apr 25, 2012 at 2:47
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    $\begingroup$ Oh dear. You cannot add $1$ and $\dfrac{1}{\cos ^2 x}$ without getting a common denominator! It's close to D and E, but neither of those. Can you see why it's close? $\endgroup$ Apr 25, 2012 at 2:48
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    $\begingroup$ Guess the socratic method loses to spoonfeeding... $\endgroup$ Apr 25, 2012 at 2:49
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    $\begingroup$ @TheChaz nice :D $\endgroup$
    – Vadim
    Apr 25, 2012 at 2:57
  • $\begingroup$ Test of Math as a Foreign Language? $\endgroup$
    – Vadim
    Apr 25, 2012 at 3:02

3 Answers 3

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Generally, what I usually do in such multiple choice questions (if I don't know how to advance) is work backwards from the given options and see which one is consistent with the question. And it is no surprise that only one of the answers will turn out to be the correct one!
Do the same here.
But lets be clever and try out the options which have a $\sec^2{x}$. (But do note that this method isn't the best method since your $\sec^2{x}$ could later appear out of other trig ratios too. To save time, we go for the ones containing $\sec^2{x}$. If this doesn't work then we shall try the other options).

The options we will try out are , thus $A$, $B$ and $C$.

$A$ doesn't seen to work here because there is no simple relation between $\csc^2 x$ and $\sec^2x.$ It is inconclusive, but this most likely, isn't the answer.

$B$ Same seems to be the case here. If $\sec^2x+\sin^2x-\cos^2x=\sec^2x+1$ this would mean that $\sin^2x-\cos^2x=1$ . Which is not true.
$\sin^2x+\cos^2x=1$ is the correct identity. (which is what we have in $C$)

$C$ is very similar to $B$ but it is actually better than $B$. From the definition of a unit circle we have that $\sin^2x+\cos^2x=1$. If you look at the expression in $C$ you see a $\sin^2x+\cos^2x$ in it. So the expression can thus be reduced as $$\sec^2x+\underline{\sin^2x+\cos^2x}=\sec^2x+\underline{1}$$
Thus $C$ is your correct answer.

This wasn't a very difficult problem, but I hope this answer helps you solve Multiple Choice questions better. This strategy is particularly useful when you are stuck with the question. No matter how difficult (or easy) the problem may be, it usually works (it does for me!)

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$${}{}{}{}{}C{}{}{}{}{}{}{}{}{}{}{}$$

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    $\begingroup$ The centering really makes it. $\endgroup$ Apr 25, 2012 at 3:01
  • $\begingroup$ @Austin: Double dollars signs! Maybe I should have colored it... $\endgroup$ Apr 25, 2012 at 3:01
  • $\begingroup$ @TheChaz Can you also add some cool animation? $\endgroup$
    – Vadim
    Apr 25, 2012 at 3:03
  • $\begingroup$ @Vadim: As you wish $\endgroup$ Apr 25, 2012 at 3:10
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    $\begingroup$ @Dan: My inspiration $\endgroup$ Apr 25, 2012 at 12:47
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it's very easy question answer is $C$

$1+\sec^2(x)=sin^2(x)+cos^2(x)+sec^2(x)$

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1=sin^2(x)+cos^2(x) $

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    $\begingroup$ On this forum, we try to avoid language that could be construed as insulting (e.g. "it's very easy"). $\endgroup$ Apr 25, 2012 at 7:14

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