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anonymous

  • one year ago

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.

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  1. misty1212
    • one year ago
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    HI!!

  2. misty1212
    • one year ago
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    lets see if we can write this in math ok?

  3. misty1212
    • one year ago
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    \[\frac{\cos(x)}{1+\sin(x)} +\frac{1+\sin(x)}{\cos(x)}\] right ?

  4. anonymous
    • one year ago
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    \[\frac{ cosx }{ 1+sinx }+\frac{ 1+sinx }{ cosx }= 2\sec x\]

  5. misty1212
    • one year ago
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    yeah must be , since the answer is \(2\sec(x)\) lets do the addition and see it

  6. anonymous
    • one year ago
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    Isn't it 2sec x

  7. misty1212
    • one year ago
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    yes it is

  8. anonymous
    • one year ago
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    because everything else cancels out. yay

  9. misty1212
    • one year ago
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    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 \[\frac{a^2+(1+b)^2}{(1+b)a}\]

  10. misty1212
    • one year ago
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    nothing cancels yet, we gotta multiply out up top \[\frac{a^2+1+2b+b^2}{(1+b)a}\]

  11. misty1212
    • one year ago
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    then since \(a^2+b^2=1\) you have \[\frac{2+2b}{(a+b)a}=\frac{2(1+b)}{(1+b)a}=\frac{2}{a}\]

  12. misty1212
    • one year ago
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    replacing \(a\) by \(\cos(x)\) you are left with \[\frac{2}{\cos(x)}=2\sec(x)\]

  13. misty1212
    • one year ago
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    hope all steps are clear, it is easier to write with a and b instead of cosine and sine

  14. anonymous
    • one year ago
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    Thank you!! @misty1212

  15. anonymous
    • one year ago
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    @misty1212 when you are done can you help me??

  16. misty1212
    • one year ago
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    \[\color\magenta\heartsuit\]

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