Why walk $-sec(x)$ become positive $1/cos(x)$?I thought that the an unfavorable would carry and it would be $-1/cos(x)$.

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You space right. Just by looking in ~ the an interpretation of $sec(x)$, one can plainly see that $$sec (x) = frac1cos (x)$$

But don"t just take my word because that it. Your equation itself will certainly tell friend if it"s appropriate or wrong.

$$ -sec (x) = frac1cos (x) implies -frac1cos (x) = frac1cos (x) implies large-1 =1 implies extFalse$$Now, over there is a result similar come this which is true,$$ sec (-x) = frac1cos(-x) = frac1cos(x)$$

Why did the happen? due to the fact that $ cos(-x) = cos(x)$.

Again don"t simply take mine word for it. I have the right to prove it making use of the identification $cos(A pm B) = cos(A) imescos(B) mp sin(A) imessin(B)$.$$cos(-x) = cos (0-x) = cos 0 imescos x + sin 0 imessin x = 1 imescos x + 0 imessin x = cos x$$Never confuse $sec(-x)$ through $-sec(x)$. Constantly remember that$$ sec(-x) = +sec(x) quadforall xin ememberingsomer.combbR $$

$ddotsmile$ hope this help


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edited Mar 3 "14 in ~ 22:42
answer Mar 3 "14 in ~ 22:35
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NickNick
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