If tan y = cot left(frac{pi}{4} - xright), th | vedclass

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Question

(C) Given $	an y = \cot \left(\frac{\pi}{4} - xight)$.Differentiating both sides with respect to $x$:$\sec^2 y \frac{dy}{dx} = -\operatorname{cosec

Vedclass · Accepted Answer

$\frac{\operatorname{cosec}^2\left(\frac{\pi}{4}-xight)}{1+	an ^2\left(\frac{\pi}{4}-xight)}$

Vedclass · Answer

(C) Given $	an y = \cot \left(\frac{\pi}{4} - xight)$.Differentiating both sides with respect to $x$:$\sec^2 y \frac{dy}{dx} = -\operatorname{cosec}^2 \left(\frac{\pi}{4} - xight) \cdot (-1)$$\sec^2 y \frac{dy}{dx} = \operatorname{cosec}^2 \left(\frac{\pi}{4} - xight)$$\frac{dy}{dx} = \frac{\operatorname{cosec}^2 \left(\frac{\pi}{4} - xight)}{\sec^2 y}$Since $\sec^2 y = 1 + 	an^2 y$ and $	an y = \cot \left(\frac{\pi}{4} - xight)$:$\frac{dy}{dx} = \frac{\operatorname{cosec}^2 \left(\frac{\pi}{4} - xight)}{1 + 	an^2 \left(\frac{\pi}{4} - xight)}$

If $\tan y = \cot \left(\frac{\pi}{4} - x\right)$, then $\frac{dy}{dx} = $

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