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

q x^{3} - p x^{2} + (1 + q) x - p = 0

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p x^{3} - q x^{2} + (1 + p) x - q = 0

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(1 + q) x^{3} - p x^{2} + q x - p = 0

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None of these

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Solution

The correct option is C $q x^{3} - p x^{2} + (1 + q) x - p = 0$

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Q. The product of the sine of the angles of a triangle is p and the product of their cosines is q. then prove that the tangents of the angles are the roots of the equation

q x^{3} - p x^{2} + (1 + q) x - p = 0

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Q. 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

The number of points (p, q) such that $p, q ϵ {1, 2, 3, 4}$ and the equation $p x^{2} + q x + 1 = 0$ has real roots is

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

The correct option is C qx3−px2+(1+q)x−p=0

The correct option is C $q x^{3} - p x^{2} + (1 + q) x - p = 0$