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Functions of the form 1sin⁡x occur so frequently that we give them their own names:

1cos⁡x=sec⁡x1sin⁡x=\cosecx1tan⁡x=cot⁡x

Starting with the identity sin2⁡x+cos2⁡x=1, we can show that

tan2⁡x+1=sec2⁡x

and

1+cot2⁡x=\cosec2x
  1. Sketch the curve

    y=\cosecx,−2π≤x≤2π
  2. Show that

    1+cot2⁡x=\cosec2x
  3. Solve the equation

    4cot2⁡θ+7=8\cosecθ

    giving all θ satisfying 0≤θ≤2π