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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions1 Mark
MCQ
Choose the correct answer from the given four options.If $|\text{x}|\leq1,$ then $2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ is equal to:
  • ✓
    $4\tan^{-1}\text{x}$
  • B
    $0$
  • C
    $\frac{\pi}{2}$
  • D
    $\pi$
✓

Answer

Correct option: A.
$4\tan^{-1}\text{x}$
We have, $2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
Let $\text{x}=\tan\theta$
$\therefore\ 2\tan^{-1}\tan\theta+\sin^{-1}\frac{2\tan\theta}{1+\tan^2\theta}$
$[\because\ \tan^{-1}(\tan\text{x})=\text{x}]$
$=2\theta+\sin^{-1}\sin2\theta$
$\Big[\because\ \sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}\Big]$
$=2\theta+2\theta$
$[\because\ \sin^{-1}=(\sin\text{x})=\text{x}]$
$=4\theta\ [\because\ \theta=\tan^{-1}\text{x}]$
$=4\tan^{-1}\text{x}$

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