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Mathematics 22 Online
OpenStudy (anonymous):

Verify the identity. cosine of x divided by quantity one plus sine of x plus quantity one plus sine of x divided by cosine of x equals two times secant of x.

OpenStudy (misty1212):

HI!!

OpenStudy (misty1212):

lets see if we can write this in math ok?

OpenStudy (misty1212):

\[\frac{\cos(x)}{1+\sin(x)} +\frac{1+\sin(x)}{\cos(x)}\] right ?

OpenStudy (anonymous):

\[\frac{ cosx }{ 1+sinx }+\frac{ 1+sinx }{ cosx }= 2\sec x\]

OpenStudy (misty1212):

yeah must be , since the answer is \(2\sec(x)\) lets do the addition and see it

OpenStudy (anonymous):

Isn't it 2sec x

OpenStudy (misty1212):

yes it is

OpenStudy (anonymous):

because everything else cancels out. yay

OpenStudy (misty1212):

lets do the addition and see it trick i learned via @satellite73 put \(\cos(x)=a\), and \(\sin(x)=b\) get \[\frac{a}{1+b}+\frac{1+b}{a}\] when you add you get \[\frac{a^2+(1+b)^2}{(1+b)a}\]

OpenStudy (misty1212):

nothing cancels yet, we gotta multiply out up top \[\frac{a^2+1+2b+b^2}{(1+b)a}\]

OpenStudy (misty1212):

then since \(a^2+b^2=1\) you have \[\frac{2+2b}{(a+b)a}=\frac{2(1+b)}{(1+b)a}=\frac{2}{a}\]

OpenStudy (misty1212):

replacing \(a\) by \(\cos(x)\) you are left with \[\frac{2}{\cos(x)}=2\sec(x)\]

OpenStudy (misty1212):

hope all steps are clear, it is easier to write with a and b instead of cosine and sine

OpenStudy (anonymous):

Thank you!! @misty1212

OpenStudy (anonymous):

@misty1212 when you are done can you help me??

OpenStudy (misty1212):

\[\color\magenta\heartsuit\]

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