lf A, B, C are the angles of a given triangle ABC. If cosA cosB ,cos⁡C=3−18 and sinA sinB sin⁡C=3+38 ,thenThe value of tanA+ tanB +tanC isThe value of tanA tanB tanB tanC+tanC tanA isThe value of tanA, tanB and tanC are

cos⁡Acos⁡Bcos⁡C=3−18,sin⁡Asin⁡Bsin⁡C=3+38⇒tan⁡Atan⁡Btan⁡C=3+33−1----iFor a triangle,tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C=3+33−1----iiNow,A+B+C=π⇒cos⁡(A+B+C)=cos⁡π ⇒cos⁡Acos⁡Bcos⁡C(1−tan⁡Atan⁡B−tan⁡Btan⁡C−tan⁡Ctan⁡A) ⇒tan⁡Atan⁡B+tan⁡Btan⁡C+tan⁡Ctan⁡A=5+43----iii From(i ), (ii) and (iii)tanA, tanB, tanC are the roots of the equationx3−3+33−1x2+(5+43)x−(3+23)=0----ivx = 1 is the root of above equation so,  (x−1){x2−(2+23)x+(3+23)}=0 (x−1)(x−3)(x−(2+3))=0 x=1,3,2+3tan⁡A=1,tan⁡B=3,tan⁡C=2+3Therefore, A=45∘,B=60∘,C=75∘cos⁡Acos⁡Bcos⁡C=3−18,sin⁡Asin⁡Bsin⁡C=3+38⇒tan⁡Atan⁡Btan⁡C=3+33−1----iFor a triangle,tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C=3+33−1----iiNow,A+B+C=π⇒cos⁡(A+B+C)=cos⁡π ⇒cos⁡Acos⁡Bcos⁡C(1−tan⁡Atan⁡B−tan⁡Btan⁡C−tan⁡Ctan⁡A) ⇒tan⁡Atan⁡B+tan⁡Btan⁡C+tan⁡Ctan⁡A=5+43----iii From(i ), (ii) and (iii)tanA, tanB, tanC are the roots of the equationx3−3+33−1x2+(5+43)x−(3+23)=0----ivx = 1 is the root of above equation so,  (x−1){x2−(2+23)x+(3+23)}=0 (x−1)(x−3)(x−(2+3))=0 x=1,3,2+3tan⁡A=1,tan⁡B=3,tan⁡C=2+3Therefore, A=45∘,B=60∘,C=75∘cos⁡Acos⁡Bcos⁡C=3−18,sin⁡Asin⁡Bsin⁡C=3+38⇒tan⁡Atan⁡Btan⁡C=3+33−1----iFor a triangle,tan⁡A+tan⁡B+tan⁡C=tan⁡Atan⁡Btan⁡C=3+33−1----iiNow,A+B+C=π⇒cos⁡(A+B+C)=cos⁡π ⇒cos⁡Acos⁡Bcos⁡C(1−tan⁡Atan⁡B−tan⁡Btan⁡C−tan⁡Ctan⁡A) ⇒tan⁡Atan⁡B+tan⁡Btan⁡C+tan⁡Ctan⁡A=5+43----iii From(i ), (ii) and (iii)tanA, tanB, tanC are the roots of the equationx3−3+33−1x2+(5+43)x−(3+23)=0----ivx = 1 is the root of above equation so,  (x−1){x2−(2+23)x+(3+23)}=0 (x−1)(x−3)(x−(2+3))=0 x=1,3,2+3tan⁡A=1,tan⁡B=3,tan⁡C=2+3Therefore, A=45∘,B=60∘,C=75∘