First slide

Properties of ITF

Question

If $x \in (- \infty, - 1)$ then $\sin^{- 1} (\frac{2 x}{1 + x^{2}})$ equals

Moderate

A

$2 \tan^{- 1} x$
B

$π - 2 \tan^{- 1} x$
C

$- π - 2 \tan^{- 1} x$
D

none of these

Solution

Let $\tan^{- 1} x = θ .$ Then, $x = \tan θ$
Now,
$- \infty < x < - 1 \Rightarrow - \infty < \tan θ < - 1 \Rightarrow - \frac{π}{2} < θ < - \frac{π}{4}$
$\begin{array}{l} ∴ \sin^{- 1} (\frac{2 x}{1 + x^{2}}) \\ = \sin^{- 1} (\sin 2 θ) \\ = \sin^{- 1} (- \sin (π + 2 θ)) \\ = \sin^{- 1} (\sin (- π - 2 θ)) \end{array}$
$= - π - 2 θ [∵ - \frac{π}{2} < θ < - \frac{π}{4} \Rightarrow - \frac{π}{2} < - π - 2 θ < 0]$
$= - π - 2 \tan^{- 1} x$

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