`tan(arccos(sqrt(2)/2))` Evaluate each expression without using a calculator

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`tan(arccos(sqrt2/2))_`

`cosy=sqrt2/2`

Imagine a right triangle with the adjacent side of length `sqrt2` and the hypotenuse of length 2. Using the the pythagorean theorem `(sqrt2)^2+x^2=2^2`     , the opposite side of the triangle is  `x=sqrt2`  .

The tangent ratio is the opposite side divided by the adjacent side.

`tany=sqrt2/sqrt2=1`

Final answer:

`tan(arccos(sqrt2/2))=1`

 

 

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