AboutAbe Mantell Expertise Hello,
I am a college professor of mathematics and regularly teach all levels
from elementary mathematics through differential equations, and would
be happy to assist anyone with such questions!
Experience Over 15 years teaching at the college level.
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).
