Giải thích các bước giải:
Ta có:
$\dfrac{\cos2x}{1+\sin2x}$
$= \dfrac{\cos^2x-\sin^2x}{\cos^2x+\sin^2x+2\sin x\cos x}$
$= \dfrac{\left(\cos x-\sin x\right)\left(\cos x+\sin x\right)}{\left(\cos x+\sin x\right)^2}$
$= \dfrac{\cos x-\sin x}{\cos x+\sin x}$
$= \dfrac{\dfrac{1}{\sqrt{2}}\cdot \left(\cos x-\sin x\right)}{\dfrac{1}{\sqrt{2}}\cdot \left(\cos x+\sin x\right)}$
$= \dfrac{\cos x\cdot \dfrac{1}{\sqrt{2}}-\sin x\cdot \dfrac{1}{\sqrt{2}}}{\cos x\cdot \dfrac{1}{\sqrt{2}}+\sin x\cdot \dfrac{1}{\sqrt{2}}}$
$= \dfrac{\cos x\cdot \cos\left(\dfrac{\pi}{4}\right)-\sin x\cdot \sin\left(\dfrac{\pi}{4}\right)}{\cos x\cdot \sin\left(\dfrac{\pi}{4}\right)+\sin x\cdot \cos\left(\dfrac{\pi}{4}\right)}$
$= \dfrac{\cos\left(x+\dfrac{\pi}{4}\right)}{ \sin\left(x+\dfrac{\pi}{4}\right)}$
$=\cot\left(x+\dfrac{\pi}{4}\right)$
$\to đpcm$
