# tan(17pi/12) find two common angles that either add up to or differ by 17pi/12. Rewrite this problem as the tangent of either sum or a difference of those two angles.

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We want to express `(17pi)/12` as a sum/difference of common angles. The angles usually memorized are the quadrantal angles (`0,pi/2,pi,(3pi)/2,2pi,` etc...) and the multiples of `pi/6,pi/4,pi/3` .

There are an infinite number of possible answers.

Since the denominator is 12 we might select a sum/difference of angles with denominators 3 and 4. Then `(17pi)/12=(Api)/3+(Bpi)/4` .

Adding the fractions on the right hand side together we get `(4Api+3Bpi)/12` . Then setting the numerators equal we get `4A+3B=17` which again has an infinite number of solutions. One solution is to let A=2 and B=3.

Thus `(17pi)/12=(2pi)/3+(3pi)/4`

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**We can express `tan(17pi)/12` as `tan((2pi)/3+(3pi)/4)` **

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** There are many other answers; e.g. `(17pi)/12=(7pi)/6+pi/4` , etc...**

We can evaluate the expression since we know the tangent of common angles:

`tan(A+B)=(tanA+tanB)/(1-tanAtanB)` so `tan((2pi)/3+(3pi)/4)=(tan((2pi)/3)+tan((3pi)/4))/(1-tan((2pi)/3)tan((3pi)/4))`

`tan((2pi)/3)=-sqrt(3)`

`tan((3pi)/4)=-1`

So `tan((2pi)/3+(3pi)/4)=(-sqrt(3)-1)/(1-(-sqrt(3))(-1))`

`=(-1-sqrt(3))/(1-sqrt(3))` and after rationalizing we get

`=2+sqrt(3)`