How do you write an algebraic expression that is equivalent to $sec (arctan 3 x)$ $sec(arctan3x)$ ?

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Answer 1 · 2017-04-28T18:47:22

$sec ({tan}^{- 1} (3 x)) = \sqrt{1 + 9 x^{2}}$ $sec(tan^-1(3x)) =sqrt(1 + 9x^2)$

Given:

$sec ({tan}^{- 1} (3 x)) =$ $sec(tan^-1(3x)) =$

Use the square root of the square:

$\sqrt{{sec}^{2} ({tan}^{- 1} (3 x))}$ $sqrt(sec^2(tan^-1(3x))$

Use the identity ${sec}^{2} (θ) = 1 + {tan}^{2} (θ)$ $sec^2(theta) = 1 + tan^2(theta)$ :

$\sqrt{1 + {tan}^{2} ({tan}^{- 1} (3 x))}$ $sqrt(1 + tan^2(tan^-1(3x))$

Use the property $tan ({tan}^{- 1} (θ)) = θ$ $tan(tan^-1(theta)) = theta$

$\sqrt{1 + {(3 x)}^{2}}$ $sqrt(1 + (3x)^2)$

Simplify a bit:

$\sqrt{1 + 9 x^{2}}$ $sqrt(1 + 9x^2)$

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