If sec (log _2 y^2) = operatorname{cosec} (lo | vedclass

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Question

(C) Given $\sec (\log _2 y^2) = \operatorname{cosec} (\log _2 x^2)$.Using the property $\log a^b = b \log a$, we have $\sec (2 \log _2 y) = \operatorn

Vedclass · Accepted Answer

$-\frac{y}{x}$

Vedclass · Answer

(C) Given $\sec (\log _2 y^2) = \operatorname{cosec} (\log _2 x^2)$.Using the property $\log a^b = b \log a$, we have $\sec (2 \log _2 y) = \operatorname{cosec} (2 \log _2 x)$.We know that $\operatorname{cosec} 	heta = \sec (\frac{\pi}{2} - 	heta)$, so $\sec (2 \log _2 y) = \sec (\frac{\pi}{2} - 2 \log _2 x)$.This implies $2 \log _2 y = \pm (\frac{\pi}{2} - 2 \log _2 x) + 2n\pi$. Taking the principal value, $2 \log _2 y + 2 \log _2 x = \frac{\pi}{2}$.$2 \log _2 (xy) = \frac{\pi}{2} \Rightarrow \log _2 (xy) = \frac{\pi}{4}$.$xy = 2^{\frac{\pi}{4}}$.Since $2^{\frac{\pi}{4}}$ is a constant, differentiating both sides with respect to $x$ gives:$x \frac{dy}{dx} + y = 0$.Therefore, $\frac{dy}{dx} = -\frac{y}{x}$.

If $\sec (\log _2 y^2) = \operatorname{cosec} (\log _2 x^2)$, then $\frac{dy}{dx} =$

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