Solving a System: No Solution
1 answer:
You might be interested in
Tan3A=tan(2A+A)
We know that , tan(x+y)=(tanx + tany)/(1 - (tanx)(tany))
tan(2A+A)=(tan2A+tanA)/(1 - (tan2A)(tanA))— (1)
We know that , tan2x=2tanx/(1 - tan^2x)
So, by substituting tan2A in (1),we get,
=[2tanA/(1 - tan^2A) + tanA]/1- (2tanA/(1 - tan^2A))(tanA)]
=[2tanA+tanA - tan^3A]/[1 - tan^2A - 2tan^2A]
=[3tanA - tan^3A]/[1-3tan^2A]
Therefore, tan3A= [3tanA - tan^3A]/[1-3tan^2A]
We know that , tan(x+y)=(tanx + tany)/(1 - (tanx)(tany))
tan(2A+A)=(tan2A+tanA)/(1 - (tan2A)(tanA))— (1)
We know that , tan2x=2tanx/(1 - tan^2x)
So, by substituting tan2A in (1),we get,
=[2tanA/(1 - tan^2A) + tanA]/1- (2tanA/(1 - tan^2A))(tanA)]
=[2tanA+tanA - tan^3A]/[1 - tan^2A - 2tan^2A]
=[3tanA - tan^3A]/[1-3tan^2A]
Therefore, tan3A= [3tanA - tan^3A]/[1-3tan^2A]