Question

evaluate tan(arccos(2/5))

Answer 1

there is a picture of an angle whose cosine is \(\frac{2}{5}\)
what you need is the "opposite side" which you find via pythagoras

Answer 2

\[\tan = \sqrt{\sec^2 x - 1} = \sqrt{\frac{1}{\cos^2 x} -1}\]

Answer 3

solve \[2^2+b^2=5^2\] for \(b\) and then your answer will be \(\frac{b}{5}\)