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Question Number 59753 by Aditya789 last updated on 14/May/19

$$\mathrm{x}=(\mathrm{a}+\mathrm{b})\mathrm{cosx}-\mathrm{bcos}\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{b}}\right)\mathrm{x}$$$$\mathrm{y}=(\mathrm{a}+\mathrm{b})\mathrm{sinx}-\mathrm{bsin}\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{b}}\right)\mathrm{x}$$$$\mathrm{find}\frac{\mathrm{dy}}{\mathrm{dx}}=\tan (\frac{\mathrm{a}}{2\mathrm{b}}+1)\mathrm{x}$$

Answered by tanmay last updated on 14/May/19

$$x+y=(a+b)(sinx+cosx)-b[sin\left(\frac{a+b}{b}\right)x+cos\left(\frac{a+b}{b}\right)x]$$$$1+\frac{dy}{dx}=(a+b)(cosx-sinx)-b[cos\left(\frac{a+b}{b}\right)x-sin\left(\frac{a+b}{b}\right)x]\times \frac{a+b}{b}$$$$1+\frac{dy}{dx}=(a+b)(cosx-sinx)-(a+b)[cos\left(\frac{a+b}{b}\right)x-sin\left(\frac{a+b}{b}\right)x]$$$$1+\frac{dy}{dx}=(a+b)[cosx-cos\left(\frac{a+b}{b}\right)x]+(a+b)[sin\left(\frac{a+b}{b}\right)x-sinx]$$$$1+\frac{dy}{dx}=(a+b)\left[2sin\left\{\frac{x+\left(\frac{a+b}{b}\right)x}{2}\right\}sin\left\{\frac{\left(\frac{a+b}{b}\right)x-x}{2}\right\}\right]+(a+b)\left[2cos\left\{\frac{\left(\frac{a+b}{b}\right)x+x}{2}\right\}sin\left\{\frac{\left(\frac{a+b}{b}\right)x-x}{2}\right\}\right]$$$$1+\frac{dy}{dx}=(a+b)\left[2sin\left(\frac{ax+2bx}{2b}\right)sin\left(\frac{ax}{2b}\right)\right]+(a+b)\left[2cos\left(\frac{ax+2bx}{2b}\right)sin\left(\frac{ax}{2b}\right)\right]$$$$\frac{dy}{dx}+1=2(a+b)sin\left(\frac{ax}{2b}\right)[sin\left(\frac{ax+2bx}{2b}\right)+cos\left(\frac{ax+2bx}{2b}\right)]$$$$1+\frac{dy}{dx}=2(a+b)sin\left(\frac{ax}{2b}\right)[sin(\frac{ax}{2b}+x)+cos(\frac{ax}{2b}+x)]$$$$1+\frac{dy}{dx}=\frac{2(a+b)sin\left(\frac{ax}{2b}\right)cos(\frac{ax}{2b}+x)[1+tan(\frac{ax}{2b}+x)]}{}$$$$plscheckthequestion...$$

Answered by $@ty@m last updated on 14/May/19

$$Inmyopinion,thequestionshouldbe:z$$$$\mathrm{x}=(\mathrm{a}+\mathrm{b})\cos \alpha -\mathrm{bcos}\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{b}}\right)\alpha ...\left(1\right)$$$$\mathrm{y}=(\mathrm{a}+\mathrm{b})\sin \alpha -\mathrm{bsin}\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{b}}\right)\alpha ....\left(2\right)$$$$\mathrm{find}\frac{\mathrm{dy}}{\mathrm{dx}}=\tan (\frac{\mathrm{a}}{2\mathrm{b}}+1)\alpha$$$$Solution:$$$$Differentiating\left(1\right)w.r.t.\alpha$$$$\frac{dx}{d\alpha }=-(a+b)\sin \alpha +b\left(\frac{a+b}{b}\right)\sin \left(\frac{a+b}{b}\right)\alpha$$$$=(a+b)\{\sin \left(\frac{a+b}{b}\right)\alpha -\sin \alpha \}$$$$=(a+b)\{2\cos \left(\frac{a+2b}{b}\right)\alpha .\sin \frac{a}{b}\alpha \}...\left(3\right)$$$$Differentiating\left(2\right)w.r.t.\alpha$$$$\frac{dy}{d\alpha }=(a+b)\cos \alpha -b\left(\frac{a+b}{b}\right)\cos \left(\frac{a+b}{b}\right)\alpha$$$$=(a+b)\{\cos \alpha -\cos \left(\frac{a+b}{b}\right)\alpha \}$$$$=(a+b).2\sin \left(\frac{a+2b}{b}\right)\alpha \sin \frac{a}{b}\alpha ...\left(4\right)$$$$\therefore \frac{dy}{dx}=\frac{\frac{dy}{d\alpha }}{\frac{dx}{d\alpha }}$$$$=\tan \frac{a+2b}{b}\alpha$$

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