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Question Number 40276 by Raj Singh last updated on 18/Jul/18

Answered by MJS last updated on 18/Jul/18

$$\frac{d}{dx}\left[\frac{\sin x}{1+\tan x}\right]=\frac{{\cos }^{3}x-{\sin }^{3}x}{1+2\sin x\cos x}$$$$\mathrm{zeros}\mathrm{at}x=\frac{\pi }{4}(4k+1)\wedge k\in \mathbb{Z}$$$$\max \mathrm{at}x=\frac{\pi }{4}(4k+1)\wedge k=2n\wedge n\in \mathbb{Z}\Rightarrow$$$$\Rightarrow \max \mathrm{ay}x=\frac{\pi }{4}(8n+1)\wedge n\in \mathbb{Z}\wedge y=\frac{\sqrt{2}}{4}$$$$\min \mathrm{at}x=\frac{\pi }{4}(1+4k)\wedge k=2n+1\wedge n\in \mathbb{Z}\Rightarrow$$$$\Rightarrow \min \mathrm{at}x=\frac{\pi }{4}(8n+5)\wedge n\in \mathbb{Z}\wedge y=-\frac{\sqrt{2}}{4}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jul/18

$$\frac{dy}{dx}=\frac{(1+tanx)cosx-sinx\left({sec}^{2}x\right)}{{(1+tanx)}^2}$$$$formax/min\frac{dy}{dx}=0$$$$cosx+sinx-\frac{sinx}{{cos}^{2}x}=0$$$${cos}^{3}x+{sinxcos}^{2}x-sinx=0$$$${cos}^{3}x+sinx(1-{sin}^{2}x)-sinx=0$$$${cos}^{3}x-{sin}^{3}x=0$$$${tan}^{3}x=1$$$$(tanx-1)({tan}^{2}x+tanx+1)=0$$$$$$$$x=\frac{\mathrm{\Pi }}{4}$$$$\frac{dy}{dx}=\frac{(1+tanx)cosx-sinx\left({sec}^{2}x\right)}{{(1+tanx)}^2}$$$$=\frac{cosx+sinx-\frac{sinx}{{cos}^{2}x}}{{(1+tanx)}^2}$$$$=\frac{{cos}^{3}x+{sinxcos}^{2}x-sinx}{{cos}^{2}x{(1+tanx)}^2}$$$$=\frac{{cos}^{3}x+sinx(1-{sin}^{2}x)-sinx}{{cos}^{2}x{(1+tanx)}^2}$$$$=\frac{{cos}^{3}x-{sin}^{3}x}{{cos}^{2}x{(1+tanx)}^2}$$$$atx=\frac{\mathrm{\Pi }}{4}\frac{dy}{dx}=0$$$$cosx>sinxwhenx<\frac{\mathrm{\Pi }}{4}\frac{dy}{dx}=+ve$$$$cosx<sinxwhenx>\frac{\mathrm{\Pi }}{4}\frac{dy}{dx}=-ve$$$$sosignchangeof\frac{dy}{dx}from+veto-ve$$$$soatx=\frac{\mathrm{\Pi }}{4}\frac{sinx}{1+tanx}ismaximum$$$$andmaxvalueis\frac{\frac{1}{\sqrt{2}}}{1+1}=\frac{1}{2\sqrt{2}}$$$$$$

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