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INVERSE TRIGNOMETRIC FUNCTIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINVERSE TRIGNOMETRIC FUNCTIONS3 Marks
Question
Solve $\cos^{-1}\sqrt3\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}$
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Answer

$\cos^{-1}\sqrt3\text{x}+\cos^{-1}\text{x}=\frac{\pi}{2}$
$ \Rightarrow\cos^{-1}\Big[\sqrt3\text{x}\times\text{x}-\sqrt{1-\big(\sqrt3\text{x}\big)^2}\sqrt{1-\text{x}^2}\Big]=\frac{\pi}{2}$
$\Big[\because\ \cos^{-1}\text{x}+\cos^{-1}\text{y}=\cos^{-1}\Big(\text{xy}-\sqrt{1-\text{x}^2}\sqrt{1-\text{y}^2}\Big)\Big]$
$ \Rightarrow\cos^{-1}\Big[\sqrt3\text{x}^2-\sqrt{1-3\text{x}^2}\sqrt{1-\text{x}^2}\Big]=\frac{\pi}{2}$
$\Rightarrow\sqrt3\text{x}^2-\sqrt{1-3\text{x}^2}\sqrt{1-\text{x}^2}=\cos\frac{\pi}{2}$
$ \Rightarrow\sqrt3\text{x}^2=\sqrt{1-3\text{x}^2}\sqrt{1-\text{x}^2}$
$ \Rightarrow3\text{x}^2=\big(1-3\text{x}^2\big)\big(1-\text{x}^2\big)$
$ \Rightarrow3\text{x}^4=1-3\text{x}^2+3\text{x}^4-\text{x}^2$
$ \Rightarrow4\text{x}^2=1$
$ \Rightarrow\text{x}^2=\frac{1}{4}$
$ \Rightarrow\text{x}=\pm\frac{1}{2}$

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