Question 69609
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How do I go about finding the exact value
of {{{tan(pi/4-pi/3)}}}?

You have to know the formula 

{{{tan(A-B)=(tan(A)-tan(B))/(1+tan(A)tan(B))}}}

Substitute {{{pi/4}}} for A and {{{pi/3}}} for B:

{{{tan(pi/4-pi/3)=(tan(pi/4)-tan(pi/3))/(1+tan(pi/4)tan(pi/3))}}}

Now you have to know that

{{{tan(pi/4) = 1}}} and that {{{tan(pi/3)= sqrt(3)}}}

So substitute these:

{{{tan(pi/4-pi/3)=(1-sqrt(3))/(1+(1)sqrt(3))}}}

or

{{{tan(pi/4-pi/3)=(1-sqrt(3))/(1+sqrt(3))}}}

Your teacher may accept it that way, but s/he 
probably would prefer that you rationalize the 
denominator:

{{{tan(pi/4-pi/3)=((1-sqrt(3))(1-sqrt(3)))/((1+sqrt(3))(1-sqrt(3)))}}}

{{{tan(pi/4-pi/3)=(1-sqrt(3)-sqrt(3)+(sqrt(3))^2)/(1-sqrt(3)+sqrt(3)-(sqrt(3))^2)}}}

{{{tan(pi/4-pi/3)=(1-2sqrt(3)+3)/(1-3)}}}

{{{tan(pi/4-pi/3)=(4-2sqrt(3))/(-2)}}}

Factor 2 out of the numerator

{{{tan(pi/4-pi/3)=(2)(2-sqrt(3))/(-2)}}}

Cancel the {{{-2}}} in the bottom into the
{{{2}}} in the top, getting -1 in the top:

{{{tan(pi/4-pi/3)= -1(2-sqrt(3))}}}
 
Remove the parentheses:

{{{tan(pi/4-pi/3)= -2+sqrt(3)}}}

or you can save one sign by writing the
second term first

{{{tan(pi/4-pi/3)= sqrt(3)-2}}}

Edwin</pre>