Mathematics
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OpenStudy (anonymous):
Can anyone please help?
http://prntscr.com/11zddy
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OpenStudy (mertsj):
Common denominator is the product of the two denominators. Write the two fractions with a common denominator and add the numerators.
OpenStudy (anonymous):
i tried merts and i got A
OpenStudy (mertsj):
How did the cos^2 get into the numerator?
OpenStudy (mertsj):
\[\frac{1-\sin \theta}{(1-\sin \theta)(1+\sin \theta)}+\frac{1+\sin \theta}{(1-\sin \theta)(1+\sin \theta)}=\]
\[\frac{2}{1-\sin ^2\theta}=\frac{2}{\cos ^2\theta}\]
OpenStudy (anonymous):
Merts so than it will be A right
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OpenStudy (mertsj):
Do you think that these are the same:
1. \[2\cos ^2\theta \]
2. \[\frac{2}{\cos ^2\theta}\]
OpenStudy (anonymous):
no:(
OpenStudy (mertsj):
Then how could it be A?
OpenStudy (mertsj):
I have clearly posted that the result is :
\[\frac{2}{\cos ^2\theta}\]
OpenStudy (anonymous):
but it is not one of the options
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OpenStudy (mertsj):
It is one of the options. What is it equivalent to?
OpenStudy (mertsj):
What is 1/cos x ?
OpenStudy (mertsj):
\[\frac{1}{\cos x}=\]
OpenStudy (anonymous):
sec theta
OpenStudy (anonymous):
so B
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OpenStudy (mertsj):
yes
OpenStudy (anonymous):
What is the simplified form of sin(x + p)?
OpenStudy (anonymous):
-sinx or sinx
OpenStudy (mertsj):
Do you mean:
\[\sin (x+\pi)\]
OpenStudy (anonymous):
yea
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OpenStudy (anonymous):
@Mertsj
OpenStudy (anonymous):
i think it -sinx