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7.4 Q.III · Q60

Q.lim⁡x→π/4[tan⁡2x−cot⁡2xsec⁡x−cosec⁡x]\displaystyle\lim_{x\to \pi/4}\left[\frac{\tan^2x-\cot^2x}{\sec x-\operatorname{cosec}x}\right]

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tan⁡x−cot⁡x=sin⁡2x−cos⁡2xsin⁡xcos⁡x=−2cos⁡2xsin⁡2x\tan x-\cot x=\dfrac{\sin^2x-\cos^2x}{\sin x\cos x}=\dfrac{-2\cos2x}{\sin2x} and tan⁡x+cot⁡x=2sin⁡2x\tan x+\cot x=\dfrac{2}{\sin2x}, so tan⁡2x−cot⁡2x=−4cos⁡2xsin⁡22x\tan^2x-\cot^2x=\dfrac{-4\cos2x}{\sin^22x}. Also sec⁡x−csc⁡x=sin⁡x−cos⁡xsin⁡xcos⁡x=2(sin⁡x−cos⁡x)sin⁡2x\sec x-\csc x=\dfrac{\sin x-\cos x}{\sin x\cos x}=\dfrac{2(\sin x-\cos x)}{\sin2x}. Dividing: −4cos⁡2x/sin⁡22x2(sin⁡x−cos⁡x)/sin⁡2x=−2cos⁡2xsin⁡2x(sin⁡x−cos⁡x)\dfrac{-4\cos2x/\sin^22x}{2(\sin x-\cos x)/\sin2x}=\dfrac{-2\cos2x}{\sin2x(\sin x-\cos x)}. Since $\cos2x=-(\sin x-\cos x)( …

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