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Question Number 117759 by Dwaipayan Shikari last updated on 13/Oct/20

$$\frac{d\theta }{dx}=\frac{\sqrt{1-\frac{{\theta }^2}{x^2}}}{sin(\theta +x)}$$

Commented by TANMAY PANACEA last updated on 13/Oct/20

$$assuming\theta andxinradian$$$$so\theta =\alpha x$$$$\frac{d\theta }{dx}=\alpha =\frac{\sqrt{1-{\alpha }^2}}{sin(1+\alpha )x}$$$$sin(1+\alpha )x=\frac{\sqrt{1-{\alpha }^2}}{\alpha }$$$$1⩾\frac{\sqrt{1-{\alpha }^2}}{\alpha }⩾-1$$$$1⩾\frac{1-{\alpha }^2}{{\alpha }^2}⩾0$$$$2{\alpha }^2⩾1\to {\alpha }^2⩾\frac{1}{2}$$$$1-{\alpha }^2⩾0\to {\alpha }^2⩽1$$$$1⩾{\alpha }^2⩾\frac{1}{2}\to 1⩾\alpha ⩾\frac{1}{\sqrt{2}}$$$$x⩾\alpha x⩾\frac{x}{\sqrt{2}}$$$$x⩾\theta ⩾\frac{x}{\sqrt{2}}$$$$\mathit{i}\mathit{h}\mathit{a}\mathit{v}\mathit{e}\mathit{j}\mathit{u}\mathit{s}\mathit{t}\mathit{p}\mathit{u}\mathit{t}\mathit{a}\mathit{n}\mathit{i}\mathit{d}\mathit{e}\mathit{a}...$$$$$$

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