If x=2csc-1t , y=2sec-1t, t≥1 then dydx=

x=2cosec-1t,y=2sec-1t (xy)2=2cosec-1t·2sec-1t (xy)2=2cosec-1t+sec-1t (xy)2=2π2⇒xy=2π4 xdydx+y=0 dydx=-yx

If x=2csc-1t , y=2sec-1t, t≥1 then dydx=

x=2cosec-1t,y=2sec-1t (xy)2=2cosec-1t·2sec-1t (xy)2=2cosec-1t+sec-1t (xy)2=2π2⇒xy=2π4 xdydx+y=0 dydx=-yx

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If $x = \sqrt{2^{\csc^{- 1} t}}$ , y= $\sqrt{{2^{s e c}}^{- 1}} t, (|t| \geq 1)$ then $\frac{d y}{d x}$ =

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$\frac{- y}{x}$

$\frac{x}{y}$

$\frac{- x}{y}$

answer is A.

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Detailed Solution

$x = \sqrt{2^{{cosec}^{- 1} t}}, y = \sqrt{2^{\sec^{- 1} t}} {(xy)}^{2} = 2^{{cosec}^{- 1} t} \cdot 2^{\sec^{- 1} t} {(xy)}^{2} = 2^{{cosec}^{- 1} t + \sec^{- 1} t} {(xy)}^{2} = 2^{\frac{π}{2}} \Rightarrow xy = 2^{\frac{π}{4}} x \frac{dy}{dx} + y = 0 \frac{dy}{dx} = \frac{- y}{x}$