document.write( "Question 277603: Use an addition or subtraction formula to find the exact value of the expression. \r
\n" ); document.write( "\n" ); document.write( "Tan(-5pie/12)\r
\n" ); document.write( "\n" ); document.write( "The formula is tan(s-t)=tan s + tan t/ 1+tans tant
\n" ); document.write( "or
\n" ); document.write( "tan(s+t)= tan s-tan t/ 1+tans tant
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Algebra.Com's Answer #202101 by nabla(475)\"\" \"About 
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Try:
\n" ); document.write( "To make this more natural, remember that the primitive tangent function has period of pi. So we have Tan(-5pi/12) congruent to Tan(7pi/12) mod pi.\r
\n" ); document.write( "\n" ); document.write( "Now: What functions do we easily know the value to for trigonometrics? pi/3 is certainly one. So, pi/3=4pi/12---> We need to know 3pi/12=pi/4 ! This is also an easy find.\r
\n" ); document.write( "\n" ); document.write( "So, we take Tan(7pi/12)=Tan(3pi/12 + 4pi/12) Now we use your formulae:\r
\n" ); document.write( "\n" ); document.write( "Tan(3pi/12 + 4pi/12)=[Tan(pi/4)-Tan(pi/3)]/(1+Tan(pi/4)Tan(pi/3))\r
\n" ); document.write( "\n" ); document.write( "Now it remains to remember what Tan(pi/4) and Tan(pi/3) are. We use the definition of Tangent.
\n" ); document.write( "Tan(x)=Sin(x)/Cos(x).
\n" ); document.write( "So Tan(pi/4)=1.
\n" ); document.write( "Tan(pi/3)=(sqrt(3)/2)/(1/2)=sqrt(3)\r
\n" ); document.write( "\n" ); document.write( "Finally, substitute into our expression:\r
\n" ); document.write( "\n" ); document.write( "(1-sqrt(3))/(1+sqrt(3))\r
\n" ); document.write( "\n" ); document.write( "Now, simplify:\r
\n" ); document.write( "\n" ); document.write( "\"%281-sqrt%283%29%29%5E2%2F%281-3%29\"=\"%281-2%2Asqrt%283%29%2B3%29%2A%281%2F%28-2%29%29\"=\"-2-sqrt%283%29\"\r
\n" ); document.write( "\n" ); document.write( "This could be done quicker, but I hope you can see the full details of what is going on through this procedure.
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