Inverse trigonometric function

Calculus Level 2

Calculate sec ⁡ ( tan ⁡ − 1 x 3 ) . \sec\left(\tan^{-1}\frac{x}{3}\right).

x 2 + 9 x \frac{\sqrt{x^{2}+9}}{x} 3 x 2 + 9 \frac{3}{\sqrt{x^{2}+9}} x x 2 + 9 \frac{x}{\sqrt{x^{2}+9}} x 2 + 9 3 \frac{\sqrt{x^{2}+9}}{3}

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Krishna Singh by Krishna Singh , 22, India

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1 solution

Tom Engelsman
May 10, 2020

Knowing that t a n − 1 ( x 3 ) = s e c − 1 ( x 2 + 9 3 ) tan^{-1}( \frac{x}{3}) = sec^{-1}( \frac{\sqrt{x^2 + 9}}{3}) , the result is just x 2 + 9 3 . \boxed{\frac{\sqrt{x^2 + 9}}{3}}.

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