JEE Main 2021 — Mathematics Calculus

$$y(x)={\mathrm{cot}}^{-1}⁡\frac{\sqrt{1+\mathrm{sin}⁡x}+\sqrt{1-\mathrm{sin}⁡x}}{\sqrt{1+\mathrm{sin}⁡x}-\sqrt{1-\mathrm{sin}⁡x}}$$y(x)=cot−1⁡1+sin⁡x​−1−sin⁡x​1+sin⁡x​+1−sin⁡x​​

As we know that, \sqrt{1-\mathrm{sin}⁡x}=|\mathrm{sin}\frac{x}{2}-\mathrm{cos}\frac{x}{2}|&\sqrt{1+\mathrm{sin}⁡x}=|\mathrm{sin}\frac{x}{2}+\mathrm{cos}\frac{x}{2}|

$$y(x)={\mathrm{cot}}^{-1}⁡\frac{|\mathrm{cos}⁡\frac{x}{2}+\mathrm{sin}⁡\frac{x}{2}|+|\mathrm{cos}⁡\frac{x}{2}-\mathrm{sin}⁡\frac{x}{2}|}{|\mathrm{cos}⁡\frac{x}{2}+\mathrm{sin}⁡\frac{x}{2}|-|\mathrm{cos}⁡\frac{x}{2}-\mathrm{sin}⁡\frac{x}{2}|}$$y(x)=cot−1⁡∣cos⁡2x​+sin⁡2x​∣−∣cos⁡2x​−sin⁡2x​∣∣cos⁡2x​+sin⁡2x​∣+∣cos⁡2x​−sin⁡2x​∣​

$$={\mathrm{cot}}^{-1}⁡\frac{\mathrm{cos}⁡\frac{x}{2}+\mathrm{sin}⁡\frac{x}{2}+\mathrm{sin}⁡\frac{x}{2}-\mathrm{cos}⁡\frac{x}{2}}{\mathrm{cos}⁡\frac{x}{2}+\mathrm{sin}⁡\frac{x}{2}-\mathrm{sin}⁡\frac{x}{2}+\mathrm{cos}⁡\frac{x}{2}}\text{ for }x\in (\frac{\pi }{2},\pi )$$=cot−1⁡cos⁡2x​+sin⁡2x​−sin⁡2x​+cos⁡2x​cos⁡2x​+sin⁡2x​+sin⁡2x​−cos⁡2x​​ for x∈(2π​,π)

$$={\mathrm{cot}}^{-1}⁡(\frac{\mathrm{sin}⁡\frac{x}{2}}{\mathrm{cos}⁡\frac{x}{2}})$$=cot−1⁡(cos⁡2x​sin⁡2x​​)

$$={\mathrm{cot}}^{-1}⁡(\mathrm{tan}⁡\frac{x}{2})$$=cot−1⁡(tan⁡2x​)

$$={\mathrm{cot}}^{-1}⁡(\mathrm{cot}⁡(\frac{\pi }{2}-\frac{x}{2}))$$=cot−1⁡(cot⁡(2π​−2x​))

$$y(x)=\frac{\pi }{2}-\frac{x}{2}$$y(x)=2π​−2x​

$${y}^{'}(x)=-\frac{1}{2}$$y′(x)=−21​