d/dx of 1+tanx/1-tanx

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$\frac{d}{d x} [l o g | s e c x + t a n x |] =$

\int \frac{t a n x + t a n^{3} x}{1 + t a n x} d x

\frac{d}{d x} [\frac{t a n x - c o t x}{t a n x + c o t x}] =

\int \frac{d x}{2 + s i n 2 x + c o s 2 x}

If $y = {\frac{1 - t a n x}{1 + t a n x}}, s h o w t h a t \frac{d y}{d x} = \frac{- 2}{(1 + s i n 2 x)}$

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