Question

If \( x = \sqrt{2 \cosec^{-1} t} \) and \( y = \sqrt{2 \sec^{-1} t} \), \( |t| \geq 1 \), then \( \dfrac{dy}{dx} = \)

• A. \( \dfrac{x}{y} \)
• B. \( \dfrac{y}{x} \)
• C. \( \dfrac{-y}{x} \)
• D. \( \dfrac{-x}{y} \)

Hint:

When differentiating inverse trigonometric functions under a square root, always apply the chain rule carefully and simplify using known derivative formulas.

Answer 1

The Correct Option is C

We are given: \[ x = \sqrt{2 \cosec^{-1} t},
y = \sqrt{2 \sec^{-1} t} \] We need to find \( \dfrac{dy}{dx} \). First, we use the chain rule: \[ \frac{dy}{dx} = \frac{dy}{dt} . \frac{dt}{dx} \] Step 1: Differentiate \( x \) with respect to \( t \) \[ x = \sqrt{2 \cosec^{-1} t} \Rightarrow \frac{dx}{dt} = \frac{1}{2\sqrt{2 \cosec^{-1} t}} . \frac{d}{dt}(\cosec^{-1} t) \] We know: \[ \frac{d}{dt}(\cosec^{-1} t) = \frac{-1}{|t|\sqrt{t^2 - 1}} \] So: \[ \frac{dx}{dt} = \frac{-1}{2\sqrt{2 \cosec^{-1} t} . |t| \sqrt{t^2 - 1}} = \frac{-1}{2x |t| \sqrt{t^2 - 1}}
\text{(since } x = \sqrt{2 \cosec^{-1} t} \text{)} \] Step 2: Differentiate \( y \) with respect to \( t \) \[ y = \sqrt{2 \sec^{-1} t} \Rightarrow \frac{dy}{dt} = \frac{1}{2\sqrt{2 \sec^{-1} t}} . \frac{d}{dt}(\sec^{-1} t) \] We know: \[ \frac{d}{dt}(\sec^{-1} t) = \frac{1}{|t|\sqrt{t^2 - 1}} \] So: \[ \frac{dy}{dt} = \frac{1}{2\sqrt{2 \sec^{-1} t} . |t| \sqrt{t^2 - 1}} = \frac{1}{2y |t| \sqrt{t^2 - 1}}
\text{(since } y = \sqrt{2 \sec^{-1} t} \text{)} \] Step 3: Use chain rule \[ \frac{dy}{dx} = \frac{dy}{dt} . \frac{dt}{dx} = \frac{dy}{dt} . \frac{1}{\frac{dx}{dt}} = \frac{\frac{1}{2y |t| \sqrt{t^2 - 1}}}{\frac{-1}{2x |t| \sqrt{t^2 - 1}}} = \frac{1}{2y} . \frac{2x}{-1} = \frac{-x}{y} \] So: \[ \boxed{\frac{dy}{dx} = \frac{-y}{x}} \]