In-the-given-equation-below-applying-the-formula-for-the-derivative-of-inverse-trigonometric-functions-what-is-the-u-from-the-given-function-y-cosec-1-sin-1-sin-x-cos-x-

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Question Number 160933 by blackmamba last updated on 09/Dec/21

$$Inthegivenequationbelow,applying$$$$theformulaforthederivativeof$$$$inversetrigonometricfunctions,$$$$whatisthe{}^{″}u{}^{″}fromthegivenfunction.$$$$y={\mathrm{cosec}}^{-1}\left[\sin \left(\frac{1+\sin x}{\cos x}\right)\right]$$

Answered by bobhans last updated on 10/Dec/21

$$\mathrm{y}={\mathrm{cosec}}^{-1}\left[\sin \left(\frac{1+\sin \mathrm{x}}{\cos \mathrm{x}}\right)\right]$$$$\mathrm{cosec}\mathrm{y}=\sin \left(\frac{1+\sin \mathrm{x}}{\cos \mathrm{x}}\right)$$$$(-\mathrm{cosec}\mathrm{y}\cot \mathrm{y}){\mathrm{y}}^{\prime }=\left(\frac{1+\sin \mathrm{x}}{\cos {}^2\mathrm{x}}\right)\cos \left(\frac{1+\sin \mathrm{x}}{\cos \mathrm{x}}\right)$$$${\mathrm{y}}^{\prime }=\left(\frac{1+\sin \mathrm{x}}{\cos {}^2\mathrm{x}}\right)\tan \left(\frac{1+\sin \mathrm{x}}{\cos \mathrm{x}}\right)\tan \left({\mathrm{cosec}}^{-1}\left[\sin \left(\frac{1+\sin \mathrm{x}}{\cos \mathrm{x}}\right)\right]\right)$$

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