# How do you prove sec(tan^-1(sqrt3/3))?

Apr 15, 2018

$\sec \left({\tan}^{- 1} \left(\frac{\sqrt{3}}{3}\right)\right) = \frac{2 \sqrt{3}}{3}$

#### Explanation:

Let ${\tan}^{- 1} \left(\frac{\sqrt{3}}{3}\right) = A$ then $\tan A = \frac{\sqrt{3}}{3}$.

$\rightarrow \sec A = \sqrt{1 + {\tan}^{2} A} = \sqrt{1 + {\left(\frac{\sqrt{3}}{3}\right)}^{2}} = \frac{2 \sqrt{3}}{3}$

$\rightarrow A = {\sec}^{- 1} \left(\frac{2 \sqrt{3}}{3}\right)$

Now, $\sec \left({\tan}^{- 1} \left(\frac{\sqrt{3}}{3}\right)\right) = \sec \left({\sec}^{- 1} \left(\frac{2 \sqrt{3}}{3}\right)\right) = \frac{2 \sqrt{3}}{3}$

Apr 18, 2018

color(indigo)(=> 2 / sqrt 3 " or " (2 sqrt 3) / 3

#### Explanation:

$\sec \left({\tan}^{-} 1 \left(\frac{\sqrt{3}}{3}\right)\right)$

$\implies \sec \left({\tan}^{-} 1 \left(\frac{1}{\sqrt{3}}\right)\right)$

=> sec (tan^-1( tan (pi/6)), color(violet)(" as "tan (pi/6) = 1 / sqrt3 " from table above"

$\implies \sec \left(\frac{\pi}{6}\right) = \frac{1}{\cos} \left(\frac{\pi}{6}\right) = \frac{1}{\frac{\sqrt{3}}{2}}$

color(indigo)(=> 2 / sqrt 3 " or " (2 sqrt 3) / 3