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1 answer:
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]
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Answer:
x-axis
Step-by-step explanation:
The asymptote is a straight line that the curve gets closer and closer to but never touches it.
The given exponential function is .
The given graph has a horizontal asymptote,
The equation of this horizontal asymptote is y=0.
This is also refers to as the x-axis.
Therefore the asymptote is the x-axis.