Inverse Trigonometric Functions: Exam-Pattern Drill (2)

Question

Evaluate $\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3)$ and prove that it equals $\pi$.

A CBSE 12 board favourite — looks short but trips students on the principal-value sign.

Solution — Step by Step

Final answer: $\tan^{-1}1 + \tan^{-1}2 + \tan^{-1}3 = \pi$.

Why This Works

The principal value of $\tan^{-1}$ lies in $(-\pi/2, \pi/2)$. When the argument-formula gives a value outside this range, we adjust by $\pm\pi$ to land in the right branch.

For $A,B > 0$ and $AB > 1$, the sum exceeds $\pi/2$ (since both individual angles exceed $\pi/4$), so the formula needs a $+\pi$ correction. This is the rule students forget.

Alternative Method

Geometric proof. Place a $1\times 1$, $1\times 2$, and $1\times 3$ right triangle in a row. The three angles at the base sum to a straight angle (i.e., $\pi$). This visual proof is part of CBSE textbooks.

Common Mistake

Naively writing $\tan^{-1}1 + \tan^{-1}2 = \tan^{-1}(-3)$ without the $+\pi$ correction. That gives a sum of $\tan^{-1}(-3) + \tan^{-1}3 = 0$, which is clearly wrong (each individual term was positive).