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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:
$\tan^{-1}\sqrt{\frac{\text{x}}{\text{a}+\sqrt{\text{a}^2-\text{x}^2}}},-\text{a}<\text{x}<\text{a}$
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Answer

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

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