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7.4 Q.II · Q57

Q.lim⁡x→π/4[cos⁡x−sin⁡xcos⁡2x]\displaystyle\lim_{x\to \pi/4}\left[\frac{\cos x-\sin x}{\cos 2x}\right]

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cos⁡2x=cos⁡2x−sin⁡2x=(cos⁡x−sin⁡x)(cos⁡x+sin⁡x)\cos2x=\cos^2x-\sin^2x=(\cos x-\sin x)(\cos x+\sin x). So cos⁡x−sin⁡xcos⁡2x=1cos⁡x+sin⁡x\dfrac{\cos x-\sin x}{\cos2x}=\dfrac{1}{\cos x+\sin x} after cancelling (cos⁡x−sin⁡x)(\cos x-\sin x) (valid near, though not at, x=π/4x=\pi/4). At x=π/4x=\pi/4: $\cos x+\sin x=\dfrac{\sqrt2}{2}+\dfrac{\ …

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