The sum of tangents trigonometric identity gives us that
tan(α+β)=1−tanαtanβtanα+tanβ.
By letting α=tan−1x and β=tan−1y, the equivalent trigonometric identity on tan−1 is
tan−1x+tan−1y=tan−1(1−xyx+y).
Let's use this identity to calculate tan−11+tan−12+tan−13. We first have
tan−11+tan−12=tan−1(1−1×21+2)=tan−1(−3).
As such, this gives
====tan−11+tan−12+tan−13tan−1(−3)+tan−13tan−1(1−(−3)×3−3+3)tan−100
By considering the right triangles, we get that tan−11>0, tan−12>0 and tan−13>0. Hence, the sum of 3 positive terms is 0.
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Top NewestI think you missed the condition there ...... Mod(xy) should be less than one to use the formula .... Or els the RHS becomes negative where as LHS remains positive!!
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