If \(\dfrac{1-\tan x}{1+\tan x}=1-\dfrac{2\tan x}{1+\tan^2 x}\), \(0 \le x < \dfrac{\pi}{2}\), \(x \neq \dfrac{\pi}{4}\), then what is \((\sin x+\cos x)\) equal to?
1
Note that \(\dfrac{2\tan x}{1+\tan^2 x}=\sin 2x\), so the equation becomes \(\dfrac{\cos x-\sin x}{\cos x+\sin x}=1-\sin 2x=(\cos x-\sin x)^2\). Let \(u=\cos x-\sin x\) and \(v=\cos x+\sin x\); then \(\dfrac{u}{v}=u^2\), i.e. \(u(1-uv)=0\). Since \(x\neq \dfrac{\pi}{4}\), \(u\neq 0\), so \(uv=1\), i.e. \(\cos^2x-\sin^2x=\cos 2x=1\), giving \(x=0\) in the given range. Then \(\sin x+\cos x = 0+1 = 1\). The correct option is (c).
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