Q95 / 160 · TS EAMCET 2023 (12 May Shift 1) · Mathematics

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

  • A
    $\frac{\operatorname{cosec}^2\left(\frac{\pi}{4}-x\right)}{1+\cot ^2\left(\frac{\pi}{4}+x\right)}$
  • B
    $\frac{-\operatorname{cosec}^2\left(\frac{\pi}{4}-x\right)}{\sec ^2 y}$
  • C
    $\frac{\operatorname{cosec}^2\left(\frac{\pi}{4}-x\right)}{1+\tan ^2\left(\frac{\pi}{4}-x\right)}$
  • Correct answer
    D
    $\frac{\sec ^2\left(\frac{\pi}{4}+x\right)}{1+\tan ^2\left(\frac{\pi}{4}+x\right)}$

Solution

$\tan y=\cot \left(\frac{\pi}{4}-x\right)$ $\begin{aligned} \Rightarrow \quad \sec ^2 y \frac{d y}{d x}=-(-1) \operatorname{cosec}^2\left(\frac{\pi}{4}-x\right) \end{aligned}$ $\begin{aligned} \Rightarrow \quad \frac{d y}{d x} & =\frac{\operatorname{cosec}^2\left(\frac{\pi}{4}-x\right)}{\sec ^2 y} \\ & =\frac{\operatorname{cosec}^2\left(\frac{\pi}{4}-x\right)}{1+\tan ^2 y}=\frac{\operatorname{cosec}^2\left(\frac{\pi}{4}-x\right)}{1+\tan ^2\left(\frac{\pi}{4}-x\right)} . \end{aligned}$

Hence, the correct answer is option: 'D'.

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