Question

In a triangle \(\mathrm{ABC}, \mathrm{p}\) is the product of sines of the angles and \(\mathrm{q}\) is the product of their cosines. The tangents of the angles are the roots of the equation (a) \(\mathrm{qx}^{3}-\mathrm{px}^{2}+(1+\mathrm{q}) \mathrm{x}-\mathrm{p}=0\) (b) \(p x^{3}-q x^{2}+(1+q) x+q=0\) (c) \(\mathrm{qx}^{3}+\mathrm{px}^{2}+(1+\mathrm{q}) \mathrm{x}+\mathrm{p}=0\) (d) \(q x^{3}+p x^{2}-(1+q) x+p=0\)

Short answer

Answer: (a) \(qx^3-px^2+(1+q)x-p=0\)

Step-by-step solution

Convert angles to radians

The angles of triangle ABC should be summed up to 180 degrees or π radians. Let's denote the angles as A, B, and C, and convert them to radians: \(A + B + C = π\)

Determine the product of sines and cosines

Let's find the expressions for p and q. \(p = sin(A) sin(B) sin(C)\) \(q = cos(A) cos(B) cos(C)\)

Find the sum of tangents

Now, let's find the sum of the tangents of the angles A, B, and C. \(tan(A) + tan(B) + tan(C) = \frac{sin(A)}{cos(A)} + \frac{sin(B)}{cos(B)} + \frac{sin(C)}{cos(C)}\)

Simplify the sum of tangents

Multiply the three original terms of the sum by \(q\) \(tan(A)q + tan(B)q + tan(C)q = \frac{sin(A)p}{cos(A)} + \frac{sin(B)p}{cos(B)} + \frac{sin(C)p}{cos(C)}\) Observe that tan(A)q = \(\frac{sin(A)p}{cos(A)}\), tan(B)q = \(\frac{sin(B)p}{cos(B)}\), and tan(C)q = \(\frac{sin(C)p}{cos(C)}\).

Consider given equation forms

Now, let's analyze the given equations. If the sum of the tangents of the angles is a root of each equation, then the equation has to satisfy the identity obtained in step 4. (a) \(\mathrm{qx}^{3}-\mathrm{px}^{2}+(1+\mathrm{q}) \mathrm{x}-\mathrm{p}=0\) (b) \(p x^{3}-q x^{2}+(1+q) x+q=0\) (c) \(\mathrm{qx}^{3}+\mathrm{px}^{2}+(1+\mathrm{q}) \mathrm{x}+\mathrm{p}=0\) (d) \(q x^{3}+p x^{2}-(1+q) x+p=0\)

Find the correct equation

Comparing the identity from step 4 with the given equations, we can see that the equation with the same structure is: (a) \(\mathrm{qx}^{3}-\mathrm{px}^{2}+(1+\mathrm{q}) \mathrm{x}-\mathrm{p}=0\) Therefore, the correct equation for this problem is option (a).