How do you write in terms of theta tan(theta-pi/6)? Thanks.

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How do you write in terms of theta tan(theta-pi/6)? Thanks.

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Answer 1 · 2018-03-01T04:35:52

$tan (θ - \frac{π}{6}) = \frac{\sqrt{3} tan θ - 1}{\sqrt{3} + tan θ}$ $color (blue)(tan (theta-pi/6)=(sqrt3 tan theta-1)/(sqrt3+tan theta))$

Explanation:

So we have
$tan (θ - \frac{π}{6})$ $tan (theta-pi/6)$

$tan (A - B) = \frac{tan A - tan B}{1 + tan A tan B}$ $color (green)(Tan (A-B)=(tan A-tanB)/(1+tan A tanB)$

Here $A = θ, B = \frac{π}{6}$ $A=theta" ,"B=pi/6$

Plugging in values

$tan (θ - \frac{π}{6}) = \frac{tan θ - tan (\frac{π}{6})}{1 + tan θ \cdot tan (\frac{π}{6})}$ $tan (theta-pi/6)= (tan theta-tan (pi/6))/(1+tan theta*tan (pi/6))$

$tan (\frac{π}{6}) = \frac{1}{\sqrt{3}}$ $color (red)(tan (pi/6)=1/sqrt3)$

$tan (θ - \frac{π}{6}) = \frac{tan θ - \frac{1}{\sqrt{3}}}{1 + tan θ \cdot \frac{1}{\sqrt{3}}}$ $tan (theta-pi/6)= (tan theta-1/sqrt3)/(1+tan theta*1/sqrt3)$

$tan (theta-pi/6)=((sqrt3 tan theta-1)*cancel (1/sqrt3))/((sqrt3+tan theta)*cancel (1/sqrt3))$

$tan (theta-pi/6)=(sqrt3 tan theta-1)/(sqrt3+tan theta)$

Hope it helps!

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How do you write in terms of theta tan(theta-pi/6)? Thanks.

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