How do you evaluate $sqrt2+tan^-1(root3(5))$ ?

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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, $sqrt2+tan^(-1)(5^(1/3))$

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

$=1.4142+(59.681/180 xx pi)$

$=1.4142+1.0416=2.4558$ , nearly.

How do you evaluate sqrt2+tan^-1(root3(5))sqrt2+tan^-1(root3(5))?