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Exercise 11.5 · Q6

Q.cos⁡2xsin⁡2xcos⁡2x\dfrac{\cos 2x}{\sin^2 x\cos^2 x}

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Expanding cos⁡2x\cos2x in terms of sin⁡2x,cos⁡2x\sin^2x,\cos^2x lets the fraction split into two standard derivatives.

Step 1. Expand the numerator. cos⁡2x=cos⁡2x−sin⁡2x\cos2x=\cos^2x-\sin^2x.

Step 2. Split the fraction. cos⁡2x−sin⁡2xsin⁡2xcos⁡2x=1sin⁡2x−1cos⁡2x=csc⁡2x−sec⁡2x\dfrac{\cos^2x-\sin^2x}{\sin^2x\cos^2x}=\dfrac1{\sin^2x}-\dfrac1{\cos^2x}=\csc^2x-\sec^2x. …

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