For y=sin ^{-1}left{frac{5 x+12 sqrt{1-x^{2}} | vedclass

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(B) Given $y=\sin ^{-1}\left(\frac{5 x+12 \sqrt{1-x^{2}}}{13}ight)$.Let $x = \cos 	heta$, then $\sqrt{1-x^2} = \sin 	heta$.Also, let $\sin \alpha

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$(1,-1)$

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(B) Given $y=\sin ^{-1}\left(\frac{5 x+12 \sqrt{1-x^{2}}}{13}ight)$.Let $x = \cos 	heta$, then $\sqrt{1-x^2} = \sin 	heta$.Also, let $\sin \alpha = \frac{5}{13}$, then $\cos \alpha = \frac{12}{13}$.Substituting these into the expression for $y$:$y = \sin^{-1}(\sin \alpha \cos 	heta + \cos \alpha \sin 	heta)$$y = \sin^{-1}(\sin(\alpha + 	heta)) = \alpha + 	heta$$y = \sin^{-1}(\frac{5}{13}) + \cos^{-1}(x)$Differentiating with respect to $x$:$y_1 = \frac{dy}{dx} = 0 - \frac{1}{\sqrt{1-x^2}} = -\frac{1}{\sqrt{1-x^2}}$Multiplying by $\sqrt{1-x^2}$ gives $y_1 \sqrt{1-x^2} = -1$.Squaring both sides: $y_1^2 (1-x^2) = 1$.Differentiating again with respect to $x$:$2y_1 y_2 (1-x^2) + y_1^2 (-2x) = 0$Dividing by $2y_1$ (assuming $y_1 
eq 0$):$y_2(1-x^2) - x y_1 = 0$.Comparing this with $a(1-x^2)y_2 + bxy_1 = 0$, we get $a=1$ and $b=-1$.Thus, $(a, b) = (1, -1)$.

For $y=\sin ^{-1}\left\{\frac{5 x+12 \sqrt{1-x^{2}}}{13}\right\} ;|x| \leq 1$, if $a\left(1-x^{2}\right) y_{2}+b x y_{1}=0$ then $(a, b)=$

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