For $x = -2$, the value of $ an^{-1}\left(\dfrac{2x}{1-x^2} ight)$ e…

Question

For $x = -2$, the value of $	an^{-1}\left(\dfrac{2x}{1-x^2}ight)$ equals

Accepted Answer

B. $\pi + 2	an^{-1}(-2)$

The identity $	an^{-1}\left(	frac{2x}{1-x^2}ight) = 2	an^{-1}(x)$ only holds for $-1 \le x \le 1$.
For $x < -1$, the principal value of $	an^{-1}\left(	frac{2x}{1-x^2}ight)$ jumps by $\pi$: the correct piecewise rule gives $\pi + 2	an^{-1}(x)$.
Substituting $x = -2$: answer is $\pi + 2	an^{-1}(-2)$.
(Numerical check: $	frac{2(-2)}{1-4} = 	frac{4}{3}$, so the LHS is $	an^{-1}(4/3) \approx 0.927$; RHS is $\pi + 2(-1.107) \approx 0.927$. Match.)

Answer

$\pi - 2	an^{-1}(2)$

Answer

$-\pi + 2	an^{-1}(-2)$

Answer

$2	an^{-1}(-2)$

For $x = -2$, the value of $\tan^{-1}\left(\dfrac{2x}{1-x^2}\right)$ equals

Keep practising Unit 1A2: Trigonometric Functions & Inverse Trig