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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions5 Marks
Question
Show that $2\tan^{-1}\text{x}+\sin^{-1}\frac{2\text{x}}{1+\text{x}^2}$ is constant for $\text{x}\geq1,$ find that constant.
✓

Answer

$2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$
  1. For x > 1
$ =2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$

$=\pi-\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ $\Big[\because\ 2\tan^{-1}\text{x}=\pi-\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big),\text{x}>1\Big]$

$=\pi$
  1. For x = 1
$=2\tan^{-1}\text{x}+\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$

$=2\tan^{-1}(1)+\sin^{-1}\bigg(\frac{2(1)}{1+(1)^2}\bigg)$

$=2\tan^{-1}(1)+\sin^{-1}(1)$

$=2\Big(\frac{\pi}{4}\Big)+\frac{\pi}{2}$

$=\pi$

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