To prove: cot − 1 ⁡ 7 + cot − 1 ⁡ 8 + cot − 1 ⁡ 18 = cot − 1 ⁡ 3

Question

To prove:       cot          −      1        ⁡  7  +      cot          −      1        ⁡  8  +      cot          −      1        ⁡  18  =      cot          −      1        ⁡  3

Blaine Andrews · Accepted Answer

The statement is equivalent to  arccot  ⁡  7  +  arccot  ⁡  8  =  arccot  ⁡  3  −  arccot  ⁡  18The right-hand side is positive because the arccotangent is decreasing, so it is in the interval   (  0  ,  π      /    2  ) and we have to ensure that the same holds for the left-hand side, but this is easy, because 7&amp;gt;1 and 8&amp;gt;1, hence  arccot  ⁡  7  +  arccot  ⁡  8  &amp;lt;  2  arccot  ⁡  1  =      π    2  The two sides are equal if and only if their cotangents are.Since  cot  ⁡  (  α  +  β  )  =            cot      ⁡      α      cot      ⁡      β      −      1              cot      ⁡      α      +      cot      ⁡      β          cot  ⁡  (  α  −  β  )  =            cot      ⁡      α      cot      ⁡      β      +      1              cot      ⁡      β      −      cot      ⁡      α      this amounts to proving that            7      ⋅      8      −      1              7      +      8        =            3      ⋅      18      +      1              18      −      3

To prove: cot <mrow class="MJX-TeXAtom-ORD"> − 1

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