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If $\sin(\sin^{-1} \frac{2}{5} + \cos ^{-1} x) =1$, then $x$ equals

  1. $1$
  2. $\frac{2}{5}$
  3. $\frac{3}{5}$
  4. None of these
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$sin \left (sin^{-1}\ \frac{2}{5} + cos^{-1}\ x \right) =1$

$\implies  \left (sin^{-1}\ \frac{2}{5} + cos^{-1}\ x \right) =sin^{-1}\ 1$

$\implies  sin^{-1}\ \frac{2}{5} + cos^{-1}\ x  =\frac{\pi}{2}$

$\implies  cos^{-1}\ x  =\frac{\pi}{2}- sin^{-1}\ \frac{2}{5}$

$\implies  cos^{-1}\ x  =cos^{-1}\ \frac{2}{5}$          $( \because sin^{-1}\ \frac{2}{5} + cos^{-1}\  \frac{2}{5}=\frac{\pi}{2})$

$\implies x= \frac{2}{5}$

$\therefore$ Option $B.$ is correct

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