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Question Number 96729 by qwertasdf last updated on 04/Jun/20

$$Provethat\ln \mid \sec \left(x\right)+\tan \left(x\right)\mid ={\tanh }^{-1}\left(\sin \left(x\right)\right)$$

Commented by prakash jain last updated on 04/Jun/20

$${\tanh }^{-1}\left(x\right)=\ln {\left(\frac{1+x}{1-x}\right)}^{1/2}$$$${\tanh }^{-1}\left(\sin \left(x\right)\right)=\ln \frac{1+\sin x}{1-\sin x}$$$$\mathrm{When}\sin x\ne 1$$$$\ln {\left(\frac{1+\sin x}{1-\sin x}\times \frac{1+\sin x}{1+\sin x}\right)}^{1/2}$$$$\ln {\left(\frac{1+2\sin x+{\sin }^{2}x}{1-{\sin }^{2}x}\right)}^{1/2}$$$$=\ln {(\frac{1}{{\cos }^{2}x}+2\tan x\sec x+{\tan }^{2}x)}^{1/2}$$$$=\ln {\left({(\sec x+\tan x)}^2\right)}^{1/2}$$$$=\ln \mid \sec x+\tan x\mid$$$$\mathrm{equality}\mathrm{only}\mathrm{valid}\mathrm{when}\sin x\ne 1$$

Answered by Ar Brandon last updated on 04/Jun/20

$$\mathcal{L}\mathrm{et};\mathrm{u}=\ln \mid \mathrm{secx}+\mathrm{tanx}\mid \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{secx}$$$$\mathrm{v}={\tanh }^{-1}\left(\mathrm{sinx}\right)\Rightarrow \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{cosx}}{1-{\sin }^2\mathrm{x}}$$$$=\frac{\mathrm{cosx}}{{\cos }^2\mathrm{x}}=\mathrm{secx}$$$$\Rightarrow \mathrm{u}=\mathrm{v}$$

Commented by Ar Brandon last updated on 04/Jun/20

$$\mathrm{But}\mathrm{in}\mathrm{this}\mathrm{case}\mathrm{the}\mathrm{constants}\mathrm{are}\mathrm{all}0$$$$\mathcal{I}\mathrm{feel}\mathrm{that}\mathrm{makes}\mathrm{a}\mathrm{difference}.\mathrm{What}\mathrm{do}$$$$\mathrm{you}\mathrm{think}\mathrm{Sir}?$$

Commented by prakash jain last updated on 04/Jun/20

$$\frac{du}{dx}=\frac{dv}{dx}\nRightarrow u=v$$$$ex:$$$$u=x^2+c$$$$v=x^2$$

Commented by prakash jain last updated on 04/Jun/20

$$\mathrm{But}\mathrm{then}\mathrm{you}\mathrm{will}\mathrm{need}\mathrm{a}\mathrm{formal}$$$$\mathrm{proof}\mathrm{that}\mathrm{there}\mathrm{is}\mathrm{no}\mathrm{constant}\mathrm{term}.$$$$\mathrm{For}\mathrm{example}\mathrm{function}\mathrm{like}\cos x,e^x$$$$\mathrm{have}\mathrm{terms}\mathrm{independent}\mathrm{of}x.$$

Commented by prakash jain last updated on 04/Jun/20

$$\mathrm{Also}(\sec \left(x\right)+\tan \left(x\right))\mathrm{has}\mathrm{terms}$$$$\mathrm{independent}\mathrm{of}x.$$$$\ln \mid \sec x+\tan x\mid \mathrm{does}\mathrm{not}\mathrm{have}$$$$\mathrm{terms}\mathrm{independent}\mathrm{of}x.$$

Commented by Ar Brandon last updated on 04/Jun/20

$$\mathrm{OK}\mathrm{thanks}\mathrm{Sir}\mathrm{for}\mathrm{your}\mathrm{opinion}.\mathrm{I}\mathrm{tried}\mathrm{a}\mathrm{different}$$$$\mathrm{approach}.$$

Commented by 1549442205 last updated on 05/Jun/20

$$\mathrm{Clearly},\mathrm{by}\mathrm{your}\mathrm{argument}\mathrm{it}\mathrm{follows}\mathrm{that}\mathrm{the}\mathrm{equality}$$$$\frac{\mathrm{du}}{\mathrm{dx}}=\frac{\mathrm{dv}}{\mathrm{dx}}\mathrm{is}\mathrm{simple}\mathrm{a}\mathrm{need}\mathrm{condition}\mathrm{to}\mathrm{u}=\mathrm{v}\mathrm{but}$$$$\mathrm{it}{\mathrm{isn}}^{\prime }\mathrm{t}\mathrm{a}\mathrm{enough}\mathrm{condition}\mathrm{to}\mathrm{u}=\mathrm{v}$$

Answered by Ar Brandon last updated on 04/Jun/20

$$\mathrm{y}=\ln \mid \mathrm{secx}+\mathrm{tanx}\mid =\ln \mid \frac{1+\mathrm{sinx}}{\mathrm{cosx}}\mid =\frac{1}{2}\ln \mid \frac{{(1+\mathrm{sinx})}^2}{{\cos }^2\mathrm{x}}\mid$$$$=\frac{1}{2}\ln \mid \frac{{(1+\mathrm{sinx})}^2}{1-{\sin }^2\mathrm{x}}\mid =\frac{1}{2}\ln \mid \frac{{(1+\mathrm{sinx})}^2}{(1-\mathrm{sinx})(1+\mathrm{sinx})}\mid$$$$=\frac{1}{2}\ln \mid \frac{1+\mathrm{sinx}}{1-\mathrm{sinx}}\mid ={\tanh }^{-1}\left(\mathrm{sinx}\right)$$

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