How do you evaluate $\sqrt{2} + {tan}^{- 1} (\sqrt[3]{5})$ $sqrt2+tan^-1(root3(5))$ ?

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Answer 1 · 2017-01-18T14:46:15

2.456, nearly.

Here, inverse tan angle has to be expressed in radian measure.

So, $\sqrt{2} + {tan}^{- 1} (5^{\frac{1}{3}})$ $sqrt2+tan^(-1)(5^(1/3))$

$= 1.4142 + {tan}^{- 1} 1.7100)$ $=1.4142+tan^(-1)1.7100)$ , nearly

$= 1.4142 + (\frac{59.681}{180} \times π)$ $=1.4142+(59.681/180 xx pi)$

$= 1.4142 + 1.0416 = 2.4558$ $=1.4142+1.0416=2.4558$ , nearly.

How do you evaluate √2+tan−1(3√5)sqrt2+tan^-1(root3(5))?