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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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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try using the sum formula identity for tan
\[\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}\]
I got (tan(x)+tan(pi/6))/1-tan(x)tan(pi/6) how does that help me?

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do you know how to evaluate tan(pi/6)?
\[\tan(\frac{\pi}{6})=\frac{\sin(\frac{\pi}{6})}{\cos(\frac{\pi}{6})}=....\]
but the equation is tan(x+pi/6) so how does that work
so when you used the sum identity just a sec ago you don't see tan(pi/6)?
I'm asking you to evaluate tan(pi/6)
\[\tan(x+\frac{\pi}{6})=\frac{\tan(x)+\color{red}{\tan(\frac{\pi}{6})}}{1-\tan(x) \color{red}{\tan(\frac{\pi}{6})}}\] I'm asking you to evaluate this thing in red
ok i did that now what
what did you get for the thing in red?
the unit circle says tan(pi/6) is..
(so i got tan(x)+rt(3))/(1-tan(x)rt(3))
hmm shouldn't tan(pi/6) be 1/sqrt(3)?
\[\tan(\frac{\pi}{6})=\frac{\sin(\frac{\pi}{6})}{\cos(\frac{\pi}{6})}=\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}=\frac{1}{\sqrt{3}}\]
\[\tan(x+\frac{\pi}{6})=\frac{\tan(x)+\color{red}{\frac{1}{\sqrt{3}}}}{1-\tan(x) \color{red}{\frac{1}{\sqrt{3}}}}\]
multiply top and bottom of the big fraction by sqrt(3)
ok so how do i get rid of the extra fractions
multiply top and bottom of the big fraction by sqrt(3)
so (rt(3)tan(x))/(rt(3)-tanx)
you left a term out on top
sqrt(3)/sqrt(3) isn't 0
so the answer is a
Thx a ton for all the help!!!
\[\tan(x+\frac{\pi}{6})=\frac{\tan(x)+\color{red}{\frac{1}{\sqrt{3}}}}{1-\tan(x) \color{red}{\frac{1}{\sqrt{3}}}} \cdot \frac{\color{blue}{\sqrt{3}}}{ \color{blue}{\sqrt{3}}} \\ \tan(x+\frac{\pi}{6})= \frac{\color{blue}{\sqrt{3}} \tan(x)+\frac{\color{blue }{\sqrt{3}}}{\color{red}{\sqrt{3}}}}{\color{blue}{\sqrt{3}}-\tan(x) \frac{\color{blue}{\sqrt{3}}}{\color{red}{\sqrt{3}}}} \]
and yes you get a as a result

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