If sin−1(3x)+sin−1(4x)=π2 then 'x'=

sin−1(ax)+sin−1(bx)=π2⇒sin−1(ax)+cos−1(x2−b2x)=π2 ax=x2−b2x⇒a2=x2−b2 x=a2+b2 Where a=3, b=4 x=32+42=5

If sin−1(3x)+sin−1(4x)=π2 then 'x'=

sin−1(ax)+sin−1(bx)=π2⇒sin−1(ax)+cos−1(x2−b2x)=π2 ax=x2−b2x⇒a2=x2−b2 x=a2+b2 Where a=3, b=4 x=32+42=5

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If $\sin^{- 1} (\frac{3}{x}) + \sin^{- 1} (\frac{4}{x}) = \frac{π}{2}$ then $' x' =$

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\sin^{- 1} (\frac{a}{x}) + \sin^{- 1} (\frac{b}{x}) = \frac{π}{2} \Rightarrow \sin^{- 1} (\frac{a}{x}) + \cos^{- 1} (\frac{\sqrt{x^{2} - b^{2}}}{x}) = \frac{π}{2}

$\frac{a}{x} = \frac{\sqrt{x^{2} - b^{2}}}{x} \Rightarrow a^{2} = x^{2} - b^{2}$

$x = \sqrt{a^{2} + b^{2}}$

Where $a = 3, b = 4$

$x = \sqrt{3^{2} + 4^{2}} = 5$

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If $\sin^{- 1} (\frac{3}{x}) + \sin^{- 1} (\frac{4}{x}) = \frac{π}{2}$ then $' x' =$