If sin^{-1}(tan frac{pi}{4}) - sin^{-1}(sqrt{ | vedclass

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(B) Given equation: $\sin^{-1}(	an \frac{\pi}{4}) - \sin^{-1}(\sqrt{\frac{3}{x}}) = \frac{\pi}{6}$Since $	an \frac{\pi}{4} = 1$, we have $\sin^{-1}(

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$x^2 - x - 12 = 0$

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(B) Given equation: $\sin^{-1}(	an \frac{\pi}{4}) - \sin^{-1}(\sqrt{\frac{3}{x}}) = \frac{\pi}{6}$Since $	an \frac{\pi}{4} = 1$, we have $\sin^{-1}(1) - \sin^{-1}(\sqrt{\frac{3}{x}}) = \frac{\pi}{6}$We know $\sin^{-1}(1) = \frac{\pi}{2}$, so $\frac{\pi}{2} - \sin^{-1}(\sqrt{\frac{3}{x}}) = \frac{\pi}{6}$Rearranging gives $\sin^{-1}(\sqrt{\frac{3}{x}}) = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$Taking sine on both sides: $\sqrt{\frac{3}{x}} = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}$Squaring both sides: $\frac{3}{x} = \frac{3}{4}$Thus, $x = 4$.Checking the options for $x = 4$:$(A)$ $4^2 - 4 - 6 = 16 - 10 = 6 
eq 0$$(B)$ $4^2 - 4 - 12 = 16 - 16 = 0$$(C)$ $4^2 + 4 - 12 = 8 
eq 0$$(D)$ $4^2 + 4 - 6 = 14 
eq 0$Therefore, $x=4$ is a root of $x^2 - x - 12 = 0$.

Column $I$	Column $II$
$(A)$ If $a=1$ and $b=0$,then $(x, y)$	$(p)$ lies on the circle $x^2+y^2=1$
$(B)$ If $a=1$ and $b=1$,then $(x, y)$	$(q)$ lies on $(x^2-1)(y^2-1)=0$
$(C)$ If $a=1$ and $b=2$,then $(x, y)$	$(r)$ lies on $y=x$
$(D)$ If $a=2$ and $b=2$,then $(x, y)$	$(s)$ lies on $(4x^2-1)(y^2-1)=0$

If $\sin^{-1}(\tan \frac{\pi}{4}) - \sin^{-1}(\sqrt{\frac{3}{x}}) = \frac{\pi}{6}$, then $x$ is a root of the equation:

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