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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions5 Marks
Question
Write the following in the simplest form:
$\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}-1}{\text{x}}\Big\},\text{x}\neq0$
✓

Answer

$\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}-1}{\text{x}}\Big\},\text{x}\neq0$
Let, $\text{x}=\tan\theta$
$=\tan^{-1}\Big\{\frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta}\Big\}$
$=\tan^{-1}\Big\{\frac{\sec\theta-1}{\tan\theta}\Big\}$ $\big\{\text{Since},1+\tan^2\theta=\sec^2\theta\big\}$
$=\tan^{-1}\Big\{\frac{1-\cos\theta}{\sin\theta}\Big\}$ $\Big\{\text{Since,}\sec\theta=\frac{1}{\cos\theta},\tan\theta=\frac{\sin\theta}{\cos\theta}\Big\}$
$=\tan^{-1}\Bigg\{\frac{\frac{2\sin^2\theta}{2}}{\frac{2\sin\theta}{2}\frac{\cos\theta}{2}}\Bigg\}$
$\Big\{\text{Since},1-\cos\theta=\frac{2\sin^2\theta}{2},\sin\theta=\frac{2\sin\theta}{2}\frac{\cos\theta}{2}\Big\}$
$=\tan^{-1}\bigg\{\frac{\frac{\sin\theta}{2}}{\frac{\cos\theta}{2}}\bigg\}$
$=\tan^{-1}\Big(\frac{\tan\theta}{2}\Big)$ $\Big\{\text{Since},\frac{\sin\theta}{\cos\theta}=\tan\theta\Big\}$
$=\frac{\theta}{2}$
$=\frac{1}{2}\tan^{-1}\text{x}$ $\{\text{Since},\tan\theta=\text{x}\Rightarrow\theta=\tan^{-1}\text{x}\}$
Hence,
$\tan^{-1}\Big\{\frac{\sqrt{1+\text{x}^2}-1}{\text{x}}\Big\}=\frac{1}{2}\tan^{-1}\text{x}$

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