If x = sqrt{2^{csc^{-1} t}} and y = sqrt{2^{s | vedclass

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(B) Given $x = \sqrt{2^{\csc^{-1} t}}$ and $y = \sqrt{2^{\sec^{-1} t}}$.Squaring both sides,we get $x^2 = 2^{\csc^{-1} t}$ and $y^2 = 2^{\sec^{-1} t}$

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$-\frac{y}{x}$

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(B) Given $x = \sqrt{2^{\csc^{-1} t}}$ and $y = \sqrt{2^{\sec^{-1} t}}$.Squaring both sides,we get $x^2 = 2^{\csc^{-1} t}$ and $y^2 = 2^{\sec^{-1} t}$.Taking the product,$x^2 y^2 = 2^{\csc^{-1} t + \sec^{-1} t}$.Since $\csc^{-1} t + \sec^{-1} t = \frac{\pi}{2}$ for $|t| \ge 1$,we have $x^2 y^2 = 2^{\pi/2}$,which is a constant.Differentiating both sides with respect to $x$:$\frac{d}{dx}(x^2 y^2) = \frac{d}{dx}(2^{\pi/2})$$2x y^2 + x^2 (2y \frac{dy}{dx}) = 0$$2xy(y + x \frac{dy}{dx}) = 0$Since $x, y 
eq 0$,we have $y + x \frac{dy}{dx} = 0$.Therefore,$\frac{dy}{dx} = -\frac{y}{x}$.

If $x = \sqrt{2^{\csc^{-1} t}}$ and $y = \sqrt{2^{\sec^{-1} t}}$ for $|t| \ge 1$,then $\frac{dy}{dx}$ is equal to:

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