derivative of 1-tanx/1+tanx

Let y = \frac{1 - \tan x}{1 + \tan x} \Rightarrow y = \frac{1 - \frac{\sin x}{\cos x}}{1 + \frac{\sin x}{\cos x}} \Rightarrow y = \frac{\cos x - \sin x}{\cos x + \sin x} \Rightarrow \frac{dy}{dx} = \frac{(\cos x + \sin x) \times \frac{d}{dx} (\cos x - \sin x) - (\cos x - \sin x) \times \frac{d}{dx} (\cos x + \sin x)}{{(\cos x + \sin x)}^{2}} \Rightarrow \frac{dy}{dx} = \frac{(\cos x + \sin x) (- \sin x - \cos x) - (\cos x - \sin x) (- \sin x + \cos x)}{{(\cos x + \sin x)}^{2}} \Rightarrow \frac{dy}{dx} = \frac{- {(\cos x + \sin x)}^{2} - {(\cos x - \sin x)}^{2}}{{(\cos x + \sin x)}^{2}} \Rightarrow \frac{dy}{dx} = \frac{- (\cos^{2} x + \sin^{2} x + 2 \sin x . \cos x) - (\cos^{2} x + \sin^{2} x - 2 \sin x . \cos x)}{{(\cos x + \sin x)}^{2}} \Rightarrow \frac{dy}{dx} = \frac{- (1 + \sin 2 x) - (1 - \sin 2 x)}{{(\cos x + \sin x)}^{2}} \Rightarrow \frac{dy}{dx} = \frac{- 1 - \sin 2 x - 1 + \sin 2 x}{{(\cos x + \sin x)}^{2}} \Rightarrow \frac{dy}{dx} = - \frac{2}{{(\cos x + \sin x)}^{2}}

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