JAMB 1991 · UME · Q40

Simplify cos⁡2x(sec⁡2x+sec⁡2xtan⁡2x)\cos^2x(\sec^2x + \sec^2x\tan^2x).

Worked solution (try it first)
  1. Take out the common factor sec⁡2x\sec^2 x in the bracket: cos⁡2xsec⁡2x(1+tan⁡2x)\cos^2 x \sec^2 x(1 + \tan^2 x).
  2. cos⁡2xsec⁡2x=1\cos^2 x \sec^2 x = 1, because sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}.
  3. That leaves 1+tan⁡2x1 + \tan^2 x.
  4. By the identity 1+tan⁡2x=sec⁡2x1 + \tan^2 x = \sec^2 x, the answer is sec⁡2x\sec^2 x, option C.

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