How do you write an algebraic expression that is equivalent to #sec(arctan3x)#?

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Answer 1 · 2017-04-28T18:47:22

#sec(tan^-1(3x)) =sqrt(1 + 9x^2)#

Given:

#sec(tan^-1(3x)) =#

Use the square root of the square:

#sqrt(sec^2(tan^-1(3x))#

Use the identity #sec^2(theta) = 1 + tan^2(theta)#:

#sqrt(1 + tan^2(tan^-1(3x))#

Use the property #tan(tan^-1(theta)) = theta#

#sqrt(1 + (3x)^2)#

Simplify a bit:

#sqrt(1 + 9x^2)#