Question

If $x = \sqrt{2^{cosec^{-1} } t}$ and $y = \sqrt{2^{\sec^{-1}} t} (|t | \geq 1)$, then $\frac{dy}{dx}$ is equal to :

• A. $\frac{y}{x}$
• B. $\frac{x}{y}$
• C. $ - \frac{y}{x}$
• D. $ - \frac{x}{y}$

Answer 1

The Correct Option is C

Given: $x=\sqrt{2^{\text {cose }^{-1} t}}$ and $y=\sqrt{2^{\text {sec }^{-1} t}}(|t| \geq 1)$
Now, $\frac{d y}{d x}=\frac{d y / d t}{d x / d t}$
$=\frac{\frac{1}{2 \sqrt{2 \sec ^{-1} t}} 2^{\sec ^{-1} t} \ln \left(\frac{1}{t \sqrt{t^{2}-1}}\right)}{-\frac{1}{2 \sqrt{2 cosec^{-1} t}} 2^{\text {cosec }^{-1}} \ln \left(\frac{1}{t \sqrt{t^{2}-1}}\right)} $
$=-\frac{\sqrt{2^{ ec ^{-1} t}}}{\sqrt{2^{\text {omee }^{-1} y}}}=\frac{-y}{x}$