Question 731435
<pre>
We use the fact that 345° = 315°+30°, two special angles, and the identity

{{{tan(alpha+beta)=(tan(alpha)+tan(beta))/(1-tan(alpha)tan(beta))}}},

From our knowledge of special angles:

tan(315°) = -1 and tan(30°) = {{{1/sqrt(3)}}} = {{{sqrt(3)/3}}}

tan(345°) = tan(315°+30°) =  {{{(tan("315°")+tan("30°"))/(1-tan("315°")tan("30°"))}}} = {{{((-1)+(sqrt(3)/3))/(1-(-1)(sqrt(3)/3))}}} = {{{(-1+sqrt(3)/3)/(1+sqrt(3)/3)}}}

To simplify that, first we multiply top and bottom by 3 and distribute
to clear of fractions in the numerator and denominator:

{{{(3(-1+sqrt(3)/3))/(3(1+sqrt(3)/3))}}} = {{{(-3+sqrt(3))/(3+sqrt(3))}}}

Now we rationalize the denominator by multiplying top and bottom by 3-&#8730;<span style="text-decoration: overline">3</span>,
multiplying out and simplifying:

{{{((-3+sqrt(3))*(3-sqrt(3)))/((3+sqrt(3))*(3-sqrt(3)))}}} = {{{(-9+3sqrt(3)+3sqrt(3)-(sqrt(3))^2)/(9-3sqrt(3)+3sqrt(3)-(sqrt(3))^2)}}} = {{{(-9+6sqrt(3)-(sqrt(3))^2)/(9-3sqrt(3)+3sqrt(3)-(sqrt(3))^2)}}} = {{{(-9+6sqrt(3)-(sqrt(3))^2)/(9-cross(3sqrt(3))+cross(3sqrt(3))-(sqrt(3))^2)}}} = {{{(-9+6sqrt(3)-3)/(9-3)}}} =

{{{(-12+6sqrt(3))/6}}} = {{{(6(-2+sqrt(3)))/6}}} = {{{(cross(6)(-2+sqrt(3)))/cross(6)}}} = -2+&#8730;<span style="text-decoration: overline">3</span>

It's always a good idea to check such answers with a calculator,
getting and comparing the decimal approximations:

By calculator tan(345°) = -.2679491924

and by calculator

-2+&#8730;<span style="text-decoration: overline">3</span> = -.2679491924

Edwin</pre>