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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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A solution of $\sin^{-1}(1)-\sin^{-1}\big(\large\frac{\sqrt 3}{x^2}\big)$$-\large\frac{\pi}{6}$$=0$ is

$(A)\;\sqrt 3\qquad(B)\;\pm \sqrt 2\qquad(C)\;x=1\qquad(D)\;x=\large\frac{1}{\sqrt 2}$

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  • $sin^{-1}1=\large\frac{\pi}{2}$
The above given equation can be written as $\large\frac{\pi}{2}-\frac{\pi}{6}$$=\sin^{-1}\big(\large\frac{\sqrt 3}{x^2}\big)$
$\Rightarrow \sin\large\frac{\pi}{3}=\frac{\sqrt 3}{x^2}$
$\large\frac{\sqrt 3}{2}=\frac{\sqrt 3}{x^2}$
$x=\pm\sqrt 2$
Hence (b) is the correct answer.
answered Oct 11, 2013 by sreemathi.v
edited Mar 26, 2014 by rvidyagovindarajan_1

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