Why does $-\sec(x)$ become positive $1/\cos(x)$?I thought that the negative would carry and also it would certainly be $-1/\cos(x)$.

You are watching: Secx-1/1-cosx=secx


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You room right. Just by looking in ~ the meaning of $\sec(x)$, one can plainly see that $$\sec (x) = \frac1\cos (x)$$

But don"t simply take my word because that it. Her equation itself will certainly tell girlfriend if it"s ideal or wrong.

$$ -\sec (x) = \frac1\cos (x) \implies -\frac1\cos (x) = \frac1\cos (x) \implies \large-1 =1 \implies \textFalse$$Now, there is a result similar come this i m sorry is true,$$ \sec (-x) = \frac1\cos(-x) = \frac1\cos(x)$$

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

Again don"t simply take mine word because that it. I deserve to prove it making use of the identity $\cos(A \pm B) = \cos(A)\times\cos(B) \mp \sin(A)\times\sin(B)$.$$\cos(-x) = \cos (0-x) = \cos 0\times\cos x + \sin 0\times\sin x = 1\times\cos x + 0\times\sin x = \cos x$$Never confuse $\sec(-x)$ with $-\sec(x)$. Constantly remember that$$ \sec(-x) = +\sec(x) \quad\forall x\in\charline-picon.combbR $$

$\ddot\smile$ hope this help


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