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If $\cos\bigg(\sin^{-1}\frac{2}{5}+\cos^{-1}x\bigg)=0$,then x is equal to

$(A)\;\frac{1}{15} \\ (B)\;\frac{2}{5} \\ (C)\;0 \\ (D)\;1 $
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  • \( cos \large\frac{\pi}{2}=0\:\:or\:cos^{-1}0=\large\frac{\pi}{2}\)
  • \( sin^{-1}x+cos^{-1}x=\large\frac{\pi}{2}\)
Ans (B) \( \frac{2}{5} \)
 
Given eqn. can be written as:
\( \Rightarrow sin^{-1}\large\frac{2}{5}+cos^{-1}x=cos^{-1}0=\large\frac{\pi}{2}\)
From the above formula we can say x=\(\large\frac{2}{5}\)

 

answered Feb 18, 2013 by thanvigandhi_1
edited Mar 16, 2013 by thanvigandhi_1
 
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