How do you calculate tan⁡(arccos⁡(513)) ?

Question

Amina Hall · Accepted Answer

If you set α=arccos⁡(513)⇒cos⁡α=513 than we have to calculate:  tan⁡α=sin⁡αcos⁡α  cos⁡α=513⇒sin⁡α=1−cos2α=1−25169  =169−25169=144169=1213  So:  tan⁡α=1213513=1213⋅135=125

Aaron Hughes · Accepted Answer

tan⁡(arccos⁡(513))  Draw a triangle in the plane with vertices (513,12−(513)2),(513,0), and the origin. Then (513) is the angle between the positive x-axis and the ray beginning at the origin and passing through (513,12−(513)2). Therefore, tan⁡(arccos⁡(513)) is 12−(513)2513.  Multiply the numerator by the reciprocal of the denominator.  12−(513)2135  One to any power is one.  1−(513)2135  Apply the product rule to 513  Raise 5 to the power of 2.  1−25132135  Raise 13 to the power of 2.  1−25169135  Write 1 as a fraction with a common denomiantor.  169169−25169135  Combine the numerators over the common denominator.  169−25169135  Subtract 25 from 169.  144169135  Rewrite 144169 as 144169.  144169⋅135  Simplify the numerator.  12169⋅135  Simplify the denomiantor.  1213⋅135  Simplify terms.  125  The result can be shown in multiple forms.  Exact Form:  125  Decimal Form:  2.4  Mixed Number Form: 225

How do you calculate \tan(\arccos(\frac{5}{13})) ?

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