Question
Evaluate tan−1(1)+tan−1(2)+tan−1(3).
Solution — Step by Step
tan−1a+tan−1b=tan−1(1−aba+b) (with corrections for quadrant)
Compute tan−1(2)+tan−1(3):
1−2⋅32+3=−55=−1
Since ab=6>1 and a,b>0, we add π:
tan−1(2)+tan−1(3)=π+tan−1(−1)=π−4π=43π
tan−1(1)+43π=4π+43π=π
Final answer: π
Why This Works
The principal range of tan−1 is (−π/2,π/2). The addition formula’s RHS lies in this range only when ab<1. If a,b>0 and ab>1, the actual sum exceeds π/2, so we add π to the formula’s output.
This identity (tan−11+tan−12+tan−13=π) is a JEE classic — appears every 2-3 years. Memorise the answer if you can.
Alternative Method
Note that 1+2+3=6 and 1⋅2⋅3=6. There’s a pretty identity:
tan−1a+tan−1b+tan−1c=tan−1(1−ab−bc−caa+b+c−abc) (modulo π)
Numerator: 1+2+3−1⋅2⋅3=0. So the sum is tan−1(0)+π=π (with correction).
Common Mistake
Students forget the +π correction when ab>1. The “raw” formula gives tan−1(−1)=−π/4, which would lead to a final answer of 0. The correct sum is π — off by π.
Always check whether ab>1 when both arguments are positive. If yes, add π to the formula’s result.
For a triangle with angles A+B+C=π, the identity tanA+tanB+tanC=tanAtanBtanC holds. The numbers 1,2,3 in this question are exactly the tangents of the angles of a triangle whose angles sum to π. That’s the geometric reason the sum comes out to π.