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For every θ at which every factor is defined, (cosecθ − sinθ)(secθ − cosθ)(tanθ + cotθ) equals:

Competition practiceTrigonometry
  • Asinθ·cosθ
  • B0
  • C1
  • Dtanθ + cotθ

Step-by-Step Solution

Rewrite each bracket over a single denominator: cosecθ − sinθ = (1 − siθ)/sinθ = coθ/sinθ, secθ − cosθ = siθ/cosθ, and tanθ + cotθ = (siθ + coθ)/(sinθ·cosθ) = 1/(sinθ·cosθ). Multiplying, the first two give sinθ·cosθ, and the third is its reciprocal, so the product is 1 for every admissible θ. Check at θ = 50°: . Stopping after two brackets gives sinθ·cosθ; forgetting to invert the third bracket leaves tanθ + cotθ; and 0 would need one factor to vanish, which none does for 0° < θ < 90°.
Answer: C) 1

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