If y=(1-tanx/1+tanx) show that dy/dx=-2/1+sin2x

 

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$$\begin{gathered} \mathrm{We} \mathrm{have}: \\ \mathrm{y}=\frac{1-\mathrm{tanx}}{1+\mathrm{tanx}} \\ =\frac{1-\frac{\mathrm{sinx}}{\mathrm{cosx}}}{1+{\displaystyle \frac{\mathrm{sinx}}{\mathrm{cosx}}}} \\ \Rightarrow \mathrm{y}=\frac{\mathrm{cosx}-\mathrm{sinx}}{\mathrm{cosx}+\mathrm{sinx}} \\ \mathrm{Differentiating} \mathrm{both} \mathrm{sides} \mathrm{with} \mathrm{respect} \mathrm{to} \mathrm{x}, \mathrm{we} \mathrm{get}: \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\left(\mathrm{cosx}+\mathrm{sinx}\right){\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(\mathrm{cosx}-\mathrm{sinx}\right)-\left(\mathrm{cosx}-\mathrm{sinx}\right){\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(\mathrm{cosx}+\mathrm{sinx}\right)}{{\left(\mathrm{cosx}+\mathrm{sinx}\right)}^2} \\ =\frac{\left(\mathrm{cosx}+\mathrm{sinx}\right){\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(\mathrm{cosx}-\mathrm{sinx}\right)-\left(\mathrm{cosx}-\mathrm{sinx}\right){\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(\mathrm{cosx}+\mathrm{sinx}\right)}{{\cos }^2\mathrm{x}+{\sin }^2\mathrm{x}+1\mathrm{sinx}.\mathrm{cosx}} \\ =\frac{\left(\mathrm{cosx}+\mathrm{sinx}\right)\left(-\mathrm{sinx}-\mathrm{cosx}\right)-\left(\mathrm{cosx}-\mathrm{sinx}\right)\left(-\mathrm{sinx}+\mathrm{cosx}\right)}{{\cos }^2\mathrm{x}+{\sin }^2\mathrm{x}+2\mathrm{sinx}.\mathrm{cosx}} \\ =\frac{-{\left(\mathrm{cosx}+\mathrm{sinx}\right)}^2-{\left(\mathrm{cosx}-\mathrm{sinx}\right)}^2}{1+\sin \left(2\mathrm{x}\right)} \\ =\frac{-{\cos }^2\mathrm{x}-{\sin }^2\mathrm{x}-2\mathrm{sinx}.\mathrm{cosx}-{\cos }^2\mathrm{x}-{\sin }^2\mathrm{x}+2\mathrm{sinx}.\mathrm{cosx}}{1+\sin \left(2\mathrm{x}\right)} \\ =\frac{-2{\cos }^2\mathrm{x}-2{\sin }^2\mathrm{x}}{1+\sin \left(2\mathrm{x}\right)} \\ =\frac{-2\left({\cos }^2\mathrm{x}+{\sin }^2\mathrm{x}\right)}{1+\sin \left(2\mathrm{x}\right)}=\frac{-2}{1+\sin \left(2\mathrm{x}\right)} \\ \mathbf{\Rightarrow }\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\frac{\mathbf{-}\mathbf{2}}{\mathbf{1}\mathbf{+}\mathbf{sin}\left(\mathbf{2}\mathbf{x}\right)} \end{gathered}$$
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