Question

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

Answer 1

$y = \frac{1 - tanx}{1 + tanx} \\ y = \frac{cosx - sinx}{cosx + sinx} \\ \frac{dy}{dx} = \frac{(cosx + sinx)( - sinx - cosx) - (cosx - sinx)( - sinx + cosx)}{( {cosx + sinx})^{2} } \\ \frac{dy}{dx} = \frac{ - ( {cosx + sinx})^{2} - ( {cosx - sinx})^{2} }{( {cosx + sinx})^{2} } \\ \frac{dy}{dx} = \frac{ - 2( {cos}^{2} x + {sin}^{2} x)}{1 + 2sinxcosx} \\ \frac{dy}{dx} = \frac{ - 2}{1 + sin2x}$