Question
Evaluate tan−1(1)+tan−1(2)+tan−1(3).
Also simplify sin−1(1+x22x) for ∣x∣≤1.
Solution — Step by Step
Recall:
tan−1a+tan−1b=tan−1(1−aba+b)+kπ
where k=0 if ab<1, k=1 if a,b>0 and ab>1, and k=−1 if a,b<0 and ab>1.
For a=1,b=2: ab=2>1 and both positive, so k=1:
tan−11+tan−12=tan−1(1−21+2)+π=tan−1(−3)+π
=−tan−13+π
tan−11+tan−12+tan−13=−tan−13+π+tan−13=π
First answer: π.
Let x=tanθ where θ∈[−π/4,π/4] (since ∣x∣≤1).
1+x22x=1+tan2θ2tanθ=sin(2θ)
sin−1(1+x22x)=sin−1(sin2θ)=2θ=2tan−1x
This works because 2θ∈[−π/2,π/2], which is the principal range of sin−1.
Second answer: 2tan−1x for ∣x∣≤1.
Why This Works
The addition formula for tan−1 requires a kπ correction whenever the principal values fall outside the standard range. Most students miss this — they apply tan−1a+tan−1b=tan−1(1−aba+b) blindly and get the wrong answer.
The substitution x=tanθ converts the algebraic expression into a trigonometric identity (sin2θ), making the simplification almost automatic. This is the standard trick for 1±x22x and 1+x21−x2 patterns.
Alternative Method
For the first part, draw a triangle with legs 1,2,3 and verify geometrically that the three angles sum to π. This was Ramanujan’s favourite identity.
For the second part, differentiate both sides:
dxdsin−1(1+x22x)=1+x22
This matches the derivative of 2tan−1x, confirming the equivalence on the domain ∣x∣≤1.
Common Mistake
Students write sin−1(sin2θ)=2θ without checking whether 2θ is in the principal range. For ∣x∣>1, the formula 2tan−1x no longer holds — you need a π−2tan−1x or −π−2tan−1x correction. JEE Advanced loves to slip ∣x∣>1 values into otherwise routine problems.