If x = sqrt{a^{sin^{-1} t}} and y = sqrt{a^{c | vedclass

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(A) Given $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}}$.Squaring both sides,we get $x^2 = a^{\sin^{-1} t}$ and $y^2 = a^{\cos^{-1} t}$

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$\frac{-y}{x}$

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(A) Given $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}}$.Squaring both sides,we get $x^2 = a^{\sin^{-1} t}$ and $y^2 = a^{\cos^{-1} t}$.Taking the logarithm on both sides,we have $\log_a(x^2) = \sin^{-1} t$ and $\log_a(y^2) = \cos^{-1} t$.Adding these two equations,we get $\log_a(x^2) + \log_a(y^2) = \sin^{-1} t + \cos^{-1} t$.Using the property $\sin^{-1} t + \cos^{-1} t = \frac{\pi}{2}$,we have $\log_a(x^2 y^2) = \frac{\pi}{2}$.This implies $x^2 y^2 = a^{\pi/2}$,which is a constant.Differentiating both sides with respect to $x$,we get $\frac{d}{dx}(x^2 y^2) = \frac{d}{dx}(a^{\pi/2})$.Using the product rule,$2x y^2 + x^2 (2y \frac{dy}{dx}) = 0$.Dividing by $2xy$,we get $y + x \frac{dy}{dx} = 0$.Therefore,$\frac{dy}{dx} = \frac{-y}{x}$.

If $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}},$ then $\frac{dy}{dx} = . . . . . .$

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