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Question Number 19658 by icyfalcon999 last updated on 14/Aug/17

$$=\frac{\sec \mathrm{x}-1}{\sec \mathrm{x}+1}$$$$=\frac{\sec \mathrm{x}+(1-1)-1}{\sec \mathrm{x}+1}$$$$=\frac{(\sec \mathrm{x}+1)-2}{\sec \mathrm{x}+1}$$$$=\frac{\sec \mathrm{x}+1}{\sec \mathrm{x}+1}-\frac{2}{\sec \mathrm{x}+1}$$$$=1-\frac{2}{\sec \mathrm{x}+1}$$$$=1-2{(\sec \mathrm{x}+1)}^{-1}$$$$\frac{\mathrm{d}}{\mathrm{dx}}(1-2{(\sec \mathrm{x}+1)}^{-1})$$$$=-2(-1{(\sec \mathrm{x}+1)}^{-2})\left(\sec \mathrm{x}\tan \mathrm{x}\right)$$$$=\frac{2\sec \mathrm{x}\tan \mathrm{x}}{{(\sec \mathrm{x}+1)}^2}$$

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