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Question Number 124779 by TITA last updated on 05/Dec/20

Commented by TITA last updated on 05/Dec/20

$$pleasehelp$$

Answered by Lordose last updated on 06/Dec/20

$$\mathrm{y}={\cos }^{-1}\left(\frac{1-{\mathrm{x}}^2}{1+{\mathrm{x}}^2}\right)\mathrm{and}\mathrm{z}={\cot }^{-1}\left(\frac{1-{3\mathrm{x}}^2}{3\mathrm{x}-{\mathrm{x}}^3}\right)$$$$\mathrm{We}\mathrm{are}\mathrm{expected}\mathrm{to}\mathrm{find}\frac{\mathrm{dy}}{\mathrm{dz}}$$$$\mathrm{Set}\mathrm{x}=\tan \theta$$$$\mathrm{y}={\cos }^{-1}\left(\cos \left(2\theta \right)\right)\mathrm{and}\mathrm{z}={\cot }^{-1}\left(\cot \left(3\theta \right)\right)$$$$\mathrm{y}=2\theta \mathrm{and}\mathrm{z}=3\theta$$$$\frac{\mathrm{dy}}{\mathrm{d}\theta }=2\mathrm{and}\frac{\mathrm{dz}}{\mathrm{d}\theta }=3$$$$\frac{\mathrm{dy}}{\mathrm{dz}}=\frac{\mathrm{dy}}{\mathrm{d}\theta }\cdot \frac{1}{\mathrm{dz}/\mathrm{d}\theta }$$$$\frac{\mathrm{dy}}{\mathrm{dz}}=2\cdot \frac{1}{3}=\frac{2}{3}$$

Commented by TITA last updated on 06/Dec/20

$$thanks$$

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