This one looks straightforward — three inverse tan values added together. The trap is in handling the addition formula correctly when ab>1. Most students apply the formula mechanically and drop the π correction. Let’s do this carefully.
Solution — Step by Step
tan−1(1)=4π — this is a standard value every student should have memorised.
We now need to handle tan−1(2)+tan−1(3) separately before adding 4π.
The tan−1 function has range (−2π,2π). When we add two inverse tangents and the result should lie outside this range, the formula alone gives us the wrong quadrant. The π correction term shifts the answer into the correct range.
When a>0, b>0, and ab>1: both angles tan−1(a) and tan−1(b) are in the first quadrant, each greater than 4π. Their sum must exceed 2π, which is outside tan−1‘s range — so the formula’s raw output undershoots by exactly π.
Geometrically, think of it as the principal value “wrapping around.” The π brings it back to the actual sum.
Alternative Method
Group the first two terms instead of the last two.
For tan−1(1)+tan−1(2): ab=1×2=2>1, both positive, so add π:
Clean cancellation. Both groupings give π, which is a good internal consistency check. If you get different answers with different groupings, you’ve missed a π somewhere.
When the answer to an inverse trig sum is a “nice” multiple of π, it’s almost always because of elegant cancellation like this. If your answer has a stray tan−1 left over, re-check whether you applied the ab>1 correction.
Common Mistake
Forgetting the π correction when ab>1.
Students apply tan−1(2)+tan−1(3)=tan−1(−55)=tan−1(−1)=−4π and get a final answer of 4π+(−4π)=0.
Zero is clearly wrong — all three values tan−1(1), tan−1(2), tan−1(3) are positive, so their sum must be positive too. Always sanity-check the sign of your answer. The missing π in the formula is the single most common error in this category of JEE questions.
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