Expert: Abe Mantell - 8/14/2009

Question
Show that the equation tan(30degree+a) =2tan(60degree-a) can be written in the form (tan^2) a+ (6*(3^.5))tan a- 5=0. Many thanks

Answer
Hello from the US!  :-)

Use the sum and difference formulas for tangent:
tan(30+a)=[tan(30)+tan(a)]/[1-tan(30)tan(a)]=[sqrt(3)+3tan(a)]/[3-sqrt(3)tan(a)]
tan(60-a)=[tan(60)-tan(a)]/[1+tan(60)tan(a)]=[sqrt(3)-tan(a)]/[1+sqrt(3)tan(a)]

Substitute those into tan(30+a)=2tan(60-a) and "cross multiply" to get:
sqrt(3)+6tan(a)+3sqrt(3)tan(a)^2=6sqrt(3)-12tan(a)+2sqrt(3)tan(a)^2
Combining terms gives: sqrt(3)tan(a)^2+18tan(a)-5sqrt(3)=0
Now divide through by sqrt(3) ==> tan(a)^2+6sqrt(3)tan(a)-5=0

OK?  I hope you can follow this and fill in the details yourself.

In case you are curious to know the solutions for a, they are approximately:
.4317178425, -1.478915394, 1.662677260, -2.709874813 (plus or minus multiples of 2Pi).

Abe