Q103 / 180 · KCET 2025 · Mathematics

If $y=\frac{\cos x}{1+\sin x}$, then (a) $\frac{d y}{d x}=\frac{-1}{1+\sin x}$ (b) $\frac{d y}{d x}=\frac{1}{1+\sin x}$ (c) $\frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{1}{2} \sec ^2\left(\frac{\pi}{4}-\frac{\mathrm{x}}{2}\right)$ (d) $\frac{d y}{d x}=\frac{1}{2} \sec ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)$

  • A
    Only b is correct
  • B
    Only a is correct
  • Correct answer
    C
    Both a and c are correct
  • D
    Both $b$ and d are correct

Solution

$\begin{aligned} & y=\frac{\cos x}{1+\sin x} \\ & y^{\prime}=\frac{-\sin x(1+\sin x)-\cos x(\cos x)}{(1+\sin x)^2}=\frac{-(1+\sin x)}{(1+\sin x)^2} \\ & y^{\prime}=\frac{-1}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2}=\frac{-1}{1+\sin x} \\ & =\frac{-1}{2 \cdot\left(\frac{1}{\sqrt{2}} \cdot \cos \frac{x}{2}+\frac{1}{\sqrt{2}} \sin \frac{x}{2}\right)^2}=\frac{-1}{2 \cdot \cos ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)} \\ & =\frac{-1}{2} \cdot \sec ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)\end{aligned}$

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

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