The product of the sines of the angles of a ΔABC is ϕ and the product of their cosines is q. Then the tangent of the angles is the roots of the equation

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Solution

The correct option is A

sin A sin B sin C= $ϕ$ , cos A cos B cos C=q
And tan A tan B tan C= $\frac{ϕ}{q}$
Also tan A+tan B+tan C=tan A tan B tan C (true for any triangle)
$\Rightarrow$ tan A+tan B+tan C= $\frac{ϕ}{q}$
Now, tan A tan B+tan B tan C+tan C tan A = $\frac{s i n A s i n B c o s C + s i n B s i n C c o s A + s i n C s i n A c o s B}{c o s A c o s B c o s C}$
= $\frac{1}{q} (s i n A s i n B c o s C + s i n C (s i n B c o s A c o s B s i n A))$
= $\frac{1}{q} (s i n A s i n B c o s C + s i n C s i n (A + B))$
= $\frac{1}{q} (s i n A s i n B c o s C + s i n^{2} C)$
= $\frac{1}{q} [1 + c o s C (- c o s C + s i n A s i n B)]$
= $\frac{1}{q} [1 + c o s A c o s B c o s C] = \frac{1}{q} (1 + q)$
The equation is $x^{3} - \frac{ϕ}{q} x^{2} + \frac{1 + q}{q} x - \frac{ϕ}{q} = 0$
$\Rightarrow q x^{3} - ϕ x^{2} + (1 + q) x - ϕ = 0$

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