alleryFor $x = -2$, the value of $\tan^{-1}\left(\dfrac{2x}{1-x^2}\right)$ equals
A$\pi - 2\tan^{-1}(2)$
B$\pi + 2\tan^{-1}(-2)$
C$-\pi + 2\tan^{-1}(-2)$
D$2\tan^{-1}(-2)$
Answer & Solution
Correct answer: B. $\pi + 2\tan^{-1}(-2)$
The identity $\tan^{-1}\left(\tfrac{2x}{1-x^2}\right) = 2\tan^{-1}(x)$ only holds for $-1 \le x \le 1$.
For $x < -1$, the principal value of $\tan^{-1}\left(\tfrac{2x}{1-x^2}\right)$ jumps by $\pi$: the correct piecewise rule gives $\pi + 2\tan^{-1}(x)$.
Substituting $x = -2$: answer is $\pi + 2\tan^{-1}(-2)$.
(Numerical check: $\tfrac{2(-2)}{1-4} = \tfrac{4}{3}$, so the LHS is $\tan^{-1}(4/3) \approx 0.927$; RHS is $\pi + 2(-1.107) \approx 0.927$. Match.)
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