Let the product of the sines of the angles of a triangle is (2/3) and the product of their cosines is (1/9). If tan A , tan B and tan C are the roots of the cubic, find the sum of the products of the roots taken two at a time.

Question

Let the product of the sines of the angles of a triangle is (2/3) and the product of their cosines is (1/9). If tan A , tan B and tan C are the roots of the cubic, find the sum of the products of the roots taken two at a time. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Given sin A sin B sin C =(2/3) ; cos A cos B cos C =(1/9) ; tan A tan B tan C =(2/3) ⋅ (9/1)=6 ∴ tan A + tan B + tan C =6 [ As., in . triangle ABC , ∑ tan A = prod tan A ] Hence the cubic is x3-6 x2+(∑ tan A tan B) x2-6=0....(1) Now ∑ tan A tan B=( sin A sin B cos C+ sin B sin C cos A+ sin C sin A cos B/ cos A cos B cos C) ....(2) Now A+B+C=π cos (A+B+C)=-1 cos (A+B) cos C- sin (A+B) sin C=-1 ( cos A cos B- sin A sin B) cos C- sin C( sin A cos B+ cos A sin B)=-1 1+ cos A cos B cos C= sin A sin B cos C+ sin B sin C cos A+ sin C sin A cos B .....(3) substituting in (2), we get ∑ tan A tan B=(1+ cos A cos B cos C/ cos A cos B cos C)=(1+(1/9)/(1)9)=(10/9) ⋅ (9/1)=10 Hence the cubic is x3-6 x2+10 x-6=0 Clearly, the sum of the products of the roots taken two at a time is 10.

Q. Let the product of the sines of the angles of a triangle is 32​ and the product of their cosines is 91​. If tanA,tanB and tanC are the roots of the cubic, find the sum of the products of the roots taken two at a time.

Q. Let the product of the sines of the angles of a triangle is $\frac{2}{3}$ and the product of their cosines is $\frac{1}{9}$ . If $tan A, tan B$ and $tan C$ are the roots of the cubic, find the sum of the products of the roots taken two at a time.