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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions5 Marks
Question
Solve the following:
$\sin^{-1}\text{x}+\sin^{-1}2\text{x}=\frac{\pi}{3}$
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Answer

We know $\sin^{-1}\text{x}+\sin^{-1}\text{y}=\sin^{-1}\Big[\text{x}\sqrt{1-\text{y}^2}+\text{y}\sqrt{1-\text{x}^2}\Big]$ $\therefore\ \sin^{-1}\text{x}+\sin^{-1}2\text{x}=\frac{\pi}{3}$ $\Rightarrow\sin^{-1}\text{x}+\sin^{-1}2\text{x}=\sin^{-1}\Big(\frac{\sqrt{3}}{2}\Big)$$\Rightarrow\sin^{-1}\text{x}-\sin^{-1}\Big(\frac{\sqrt{3}}{2}\Big)=-\sin^{-1}2\text{x}$
$\Rightarrow\sin^{-1}\bigg[\text{x}\sqrt{1-\frac{3}{4}}+\frac{\sqrt3}{2}\sqrt{1-\text{x}^2}\bigg]=-\sin^{-1}2\text{x}$
$\Rightarrow\sin^{-1}\Big[\frac{\text{x}}{2}+\frac{\sqrt3}{2}\sqrt{1-\text{x}^2}\Big]=\sin^{-1}(-2\text{x})$
$\Rightarrow\frac{\text{x}}{2}+\frac{\sqrt3}{2}\sqrt{1-\text{x}^2}=-2\text{x}$
$\Rightarrow5\text{x}=-\sqrt3\sqrt{1-\text{x}^2}$
Squaring both the sides, $25\text{x}^2=3-3\text{x}^2$ $\Rightarrow28\text{x}^2=3$ $\Rightarrow\text{x}=\pm\frac{1}{2}\sqrt{\frac{3}{7}}$

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