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anonymous

  • one year ago

The equation tan(x+pi/6) is equal to _____. a. √3 tanx+1/√3-tanx b. tanx+√3/1-√3 tanx c. tanx-√3/1+√3tanx d. √3 tanx-1/√3+tanx

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  1. freckles
    • one year ago
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    try using the sum formula identity for tan

  2. freckles
    • one year ago
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    \[\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}\]

  3. anonymous
    • one year ago
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    I got (tan(x)+tan(pi/6))/1-tan(x)tan(pi/6) how does that help me?

  4. freckles
    • one year ago
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    do you know how to evaluate tan(pi/6)?

  5. freckles
    • one year ago
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    \[\tan(\frac{\pi}{6})=\frac{\sin(\frac{\pi}{6})}{\cos(\frac{\pi}{6})}=....\]

  6. anonymous
    • one year ago
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    but the equation is tan(x+pi/6) so how does that work

  7. freckles
    • one year ago
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    so when you used the sum identity just a sec ago you don't see tan(pi/6)?

  8. freckles
    • one year ago
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    I'm asking you to evaluate tan(pi/6)

  9. freckles
    • one year ago
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    \[\tan(x+\frac{\pi}{6})=\frac{\tan(x)+\color{red}{\tan(\frac{\pi}{6})}}{1-\tan(x) \color{red}{\tan(\frac{\pi}{6})}}\] I'm asking you to evaluate this thing in red

  10. anonymous
    • one year ago
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    ok i did that now what

  11. freckles
    • one year ago
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    what did you get for the thing in red?

  12. freckles
    • one year ago
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    the unit circle says tan(pi/6) is..

  13. anonymous
    • one year ago
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    (so i got tan(x)+rt(3))/(1-tan(x)rt(3))

  14. freckles
    • one year ago
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    hmm shouldn't tan(pi/6) be 1/sqrt(3)?

  15. freckles
    • one year ago
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    \[\tan(\frac{\pi}{6})=\frac{\sin(\frac{\pi}{6})}{\cos(\frac{\pi}{6})}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}\]

  16. freckles
    • one year ago
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    \[\tan(x+\frac{\pi}{6})=\frac{\tan(x)+\color{red}{\frac{1}{\sqrt{3}}}}{1-\tan(x) \color{red}{\frac{1}{\sqrt{3}}}}\]

  17. freckles
    • one year ago
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    multiply top and bottom of the big fraction by sqrt(3)

  18. anonymous
    • one year ago
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    ok so how do i get rid of the extra fractions

  19. freckles
    • one year ago
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    multiply top and bottom of the big fraction by sqrt(3)

  20. anonymous
    • one year ago
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    so (rt(3)tan(x))/(rt(3)-tanx)

  21. freckles
    • one year ago
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    you left a term out on top

  22. freckles
    • one year ago
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    sqrt(3)/sqrt(3) isn't 0

  23. anonymous
    • one year ago
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    so the answer is a

  24. anonymous
    • one year ago
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    Thx a ton for all the help!!!

  25. freckles
    • one year ago
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    \[\tan(x+\frac{\pi}{6})=\frac{\tan(x)+\color{red}{\frac{1}{\sqrt{3}}}}{1-\tan(x) \color{red}{\frac{1}{\sqrt{3}}}} \cdot \frac{\color{blue}{\sqrt{3}}}{ \color{blue}{\sqrt{3}}} \\ \tan(x+\frac{\pi}{6})= \frac{\color{blue}{\sqrt{3}} \tan(x)+\frac{\color{blue }{\sqrt{3}}}{\color{red}{\sqrt{3}}}}{\color{blue}{\sqrt{3}}-\tan(x) \frac{\color{blue}{\sqrt{3}}}{\color{red}{\sqrt{3}}}} \]

  26. freckles
    • one year ago
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    and yes you get a as a result

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