# solve this Q. y = find $\frac{dy}{dx}$.

$Given\phantom{\rule{0ex}{0ex}}y={\mathrm{tan}}^{-1}\left(\frac{\sqrt{1+\mathrm{sin}x}+\sqrt{1-\mathrm{sin}x}}{\sqrt{1+\mathrm{sin}x}-\sqrt{1-\mathrm{sin}x}}\right)\phantom{\rule{0ex}{0ex}}⇒y={\mathrm{tan}}^{-1}\left(\frac{\sqrt{1+\mathrm{sin}x}+\sqrt{1-\mathrm{sin}x}}{\sqrt{1+\mathrm{sin}x}-\sqrt{1-\mathrm{sin}x}}×\frac{\sqrt{1+\mathrm{sin}x}+\sqrt{1-\mathrm{sin}x}}{\sqrt{1+\mathrm{sin}x}+\sqrt{1-\mathrm{sin}x}}\right)\phantom{\rule{0ex}{0ex}}⇒y={\mathrm{tan}}^{-1}\left(\frac{\left(1+\mathrm{sin}x+1-\mathrm{sin}x+2\sqrt{1-{\mathrm{sin}}^{2}x}\right)}{1+\mathrm{sin}x-1+\mathrm{sin}x}\right)\phantom{\rule{0ex}{0ex}}⇒y={\mathrm{tan}}^{-1}\left(\frac{2+2\sqrt{{\mathrm{cos}}^{2}x}}{2\mathrm{sin}x}\right)\phantom{\rule{0ex}{0ex}}⇒y={\mathrm{tan}}^{-1}\left(\frac{2+2\mathrm{cos}x}{2\mathrm{sin}x}\right)\phantom{\rule{0ex}{0ex}}⇒y={\mathrm{tan}}^{-1}\left(\frac{1+\mathrm{cos}x}{\mathrm{sin}x}\right)\phantom{\rule{0ex}{0ex}}⇒y={\mathrm{tan}}^{-1}\left(\frac{2{\mathrm{cos}}^{2}\frac{x}{2}}{2\mathrm{sin}\frac{x}{2}\mathrm{cos}\frac{x}{2}}\right)\phantom{\rule{0ex}{0ex}}⇒y={\mathrm{tan}}^{-1}\left(\frac{\mathrm{cos}\frac{x}{2}}{\mathrm{sin}\frac{x}{2}}\right)\phantom{\rule{0ex}{0ex}}⇒y={\mathrm{tan}}^{-1}\left(\mathrm{co}t\frac{x}{2}\right)\phantom{\rule{0ex}{0ex}}⇒y={\mathrm{tan}}^{-1}\left(\mathrm{tan}\left(\frac{\pi }{2}-\frac{x}{2}\right)\right)\phantom{\rule{0ex}{0ex}}⇒y=\frac{\pi }{2}-\frac{x}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=0-\frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=\frac{-1}{2}$

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