Multiple choice

If $p$ is the product of the sines of angles of a triangle and $q$ the product of their cosines, the tangents of the angle are roots of the equation

  1. $q{x}^{3}-p{x}^{2}+\left(1+q\right)x-p=0$
  2. $p{x}^{3}-q{x}^{2}+\left(1+p\right)x-q=0$
  3. $\left(1+q\right){x}^{3}-p{x}^{2}+\left(1+q\right)x-p=0$

  4. none of the above
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A Correct answer
AI explanation

Let the tangents of the angles be tan A, tan B, and tan C, and let x represent a general root of the required cubic polynomial. Because A, B, and C are angles of a triangle, their sum is 180 degrees, making the sum of the tangents equal to the product of the tangents (tan A + tan B + tan C = tan A tan B tan C = p/q). The required polynomial with roots tan A, tan B, and tan C is formed by the relation x^3 - (sum of roots)x^2 + (sum of product of roots two at a time)x - (product of roots) = 0. Multiplying the entire polynomial by q and substituting q = cos A cos B cos C and p = sin A sin B sin C results in the equation q x^3 - p x^2 + (1+q)x - p = 0.

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