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If $\tan y=\cot \left(\frac{\pi}{4}-x\right)$ then $\frac{d y}{d x}=$
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Verified Answer
The correct answer is:
$\frac{\sec ^2\left(\frac{\pi}{4}+x\right)}{1+\tan ^2\left(\frac{\pi}{4}+x\right)}$
$\tan y=\cot \left(\frac{\pi}{4}-x\right)$
$\begin{aligned}
\Rightarrow \quad \sec ^2 y \frac{d y}{d x}=-(-1) \operatorname{cosec}^2\left(\frac{\pi}{4}-x\right)
\end{aligned}$
$\begin{aligned}
\Rightarrow \quad \frac{d y}{d x} & =\frac{\operatorname{cosec}^2\left(\frac{\pi}{4}-x\right)}{\sec ^2 y} \\
& =\frac{\operatorname{cosec}^2\left(\frac{\pi}{4}-x\right)}{1+\tan ^2 y}=\frac{\operatorname{cosec}^2\left(\frac{\pi}{4}-x\right)}{1+\tan ^2\left(\frac{\pi}{4}-x\right)} .
\end{aligned}$
$\begin{aligned}
\Rightarrow \quad \sec ^2 y \frac{d y}{d x}=-(-1) \operatorname{cosec}^2\left(\frac{\pi}{4}-x\right)
\end{aligned}$
$\begin{aligned}
\Rightarrow \quad \frac{d y}{d x} & =\frac{\operatorname{cosec}^2\left(\frac{\pi}{4}-x\right)}{\sec ^2 y} \\
& =\frac{\operatorname{cosec}^2\left(\frac{\pi}{4}-x\right)}{1+\tan ^2 y}=\frac{\operatorname{cosec}^2\left(\frac{\pi}{4}-x\right)}{1+\tan ^2\left(\frac{\pi}{4}-x\right)} .
\end{aligned}$
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