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$S i n^{- 1} (\frac{a}{x}) + S i n^{- 1} (\frac{b}{x}) = \frac{π}{2}$ then $x =$

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detailed solution

Correct option is B

Given sin-1ax+sin-1bx=π2⇒sin-1ax+Cos-11-bx2=π2 ∵sin-1p=cos-11-p2⇒ax=1-b2x2⇒a2x2=1-b2x2⇒x=a2+b2

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Sin−1ax+Sin−1bx=π2thenx=

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$S i n^{- 1} (\frac{a}{x}) + S i n^{- 1} (\frac{b}{x}) = \frac{π}{2}$ then $x =$