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Question Number 44622 by mondodotto@gmail.com last updated on 02/Oct/18
if y=ln[tan((𝛑/4)+(x/2))] show that  (dy/dx)=secx
$$\boldsymbol{\mathrm{if}}\:\boldsymbol{{y}}=\boldsymbol{\mathrm{ln}}\left[\boldsymbol{\mathrm{tan}}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}+\frac{\boldsymbol{{x}}}{\mathrm{2}}\right)\right]\:\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}} \\ $$$$\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}=\boldsymbol{\mathrm{sec}{x}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 02/Oct/18
(dy/dx)=(1/(tan((π/4)+(x/2))))×sec^2 ((π/4)+(x/2))×(1/2)  (dy/dx)=(1/2)×((cos((π/4)+(x/2)))/(sin((π/4)+(x/2))))×(1/(cos^2 ((π/4)+(x/2))))  (dy/dx)=(1/(2sin((π/4)+(x/2))cos((π/4)+(x/2))))=(1/(sin((π/2)+x)))=(1/(cosx))=secx
$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{{tan}\left(\frac{\pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right)}×{sec}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right)×\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{{cos}\left(\frac{\pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right)}{{sin}\left(\frac{\pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right)}×\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right)} \\ $$$$\frac{{dy}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}{sin}\left(\frac{\pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{\pi}{\mathrm{4}}+\frac{{x}}{\mathrm{2}}\right)}=\frac{\mathrm{1}}{{sin}\left(\frac{\pi}{\mathrm{2}}+{x}\right)}=\frac{\mathrm{1}}{{cosx}}={secx} \\ $$

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