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

Question

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

Infinity Learn · Accepted Answer

$$cos$$⁡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,  $$(x$$−$$1){x2$$−$$(2+23)x+(3+23)}=0$$ $$(x$$−$$1)(x$$−$$3)(x$$−$$(2+3))=0$$ $$x=1$$,$$3$$,$$2+3tan$$⁡$$A=1$$,$$tan$$⁡$$B=3$$,$$tan$$⁡$$C=2+3Therefore$$, $$A=45$$∘,$$B=60$$∘,$$C=75$$∘

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

Detailed Solution

Get Expert Academic Guidance – Connect with a Counselor Today!