The inverse tangent addition formula tan−1x+tan−1y=tan−11−xyx+y holds only when xy<1. Here x=3, y=λ, so the condition is 3λ<1, i.e. λ<31. The correct option is (C).
The formula you’ve written —
tan−13+tan−1λ=tan−1(1−3λ3+λ)
— is the standard inverse tangent addition identity, but it comes with a hidden condition. Many students apply it blindly, and that’s where mistakes happen.
Let’s understand why the condition exists.
The core idea: the range of tan−1 and the product condition
Recall that tan−1x (also written arctanx) gives an angle in (−2π,2π). So the sum of two such angles, tan−13+tan−1λ, lies in (−π,π).
The formula
tan−1x+tan−1y=tan−11−xyx+y
is derived from the tangent addition formula:
tan(A+B)=1−tanAtanBtanA+tanB
If we set A=tan−1x, B=tan−1y, then tan(A+B)=1−xyx+y.
But here’s the catch: tan−11−xyx+y always gives an angle in (−2π,2π). So the equality holds only when A+B itself lies in (−2π,2π).
When does A+B stay inside that interval? It turns out the cleanest condition is xy<1.
tan−1x+tan−1y=tan−11−xyx+yif and only ifxy<1
If xy=1, the denominator is zero — the formula breaks. If xy>1, then A+B falls outside (−2π,2π), and the right-hand side would give a different principal value (you’d need to add or subtract π).
Applying it to this problem
Here x=3 and y=λ. So the condition for the formula to be valid is:
-
Write the product condition:
xy<1⇒3λ<1
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Solve for λ:
λ<31
That’s it. No further restrictions — λ can be any real number less than 31. …