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Question Number 77123 by mathocean1 last updated on 03/Jan/20

ABC is a non−right triangle.  1) Demonstrate that  tan(A^� +B^� )=−tanC^� .  1) By using tan(A^� +B^� )=((tanA^� +tanB^� )/(1−tanA^� tanB^� ))  prove that tanA^� +tanB^� +tanC^� =tanAtanBtanC  please i need your help

$$\mathrm{ABC}\:\mathrm{is}\:\mathrm{a}\:\mathrm{non}−\mathrm{right}\:\mathrm{triangle}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{Demonstrate}\:\mathrm{that} \\ $$$$\mathrm{tan}\left(\hat {\mathrm{A}}+\hat {\mathrm{B}}\right)=−\mathrm{tan}\hat {\mathrm{C}}. \\ $$$$\left.\mathrm{1}\right)\:\mathrm{By}\:\mathrm{using}\:\mathrm{tan}\left(\hat {\mathrm{A}}+\hat {\mathrm{B}}\right)=\frac{\mathrm{tan}\hat {\mathrm{A}}+\mathrm{tan}\hat {\mathrm{B}}}{\mathrm{1}−\mathrm{tan}\hat {\mathrm{A}tan}\hat {\mathrm{B}}} \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{tan}\hat {\mathrm{A}}+\mathrm{tan}\hat {\mathrm{B}}+\mathrm{tan}\hat {\mathrm{C}}=\mathrm{tanAtanBtanC} \\ $$$$\mathrm{please}\:\mathrm{i}\:\mathrm{need}\:\mathrm{your}\:\mathrm{help} \\ $$

Answered by john santu last updated on 03/Jan/20

(1) A+B = π − C   tan (A+B) = tan (π−C)  tan (A+B) = −tan C

$$\left(\mathrm{1}\right)\:{A}+{B}\:=\:\pi\:−\:{C}\: \\ $$$$\mathrm{tan}\:\left({A}+{B}\right)\:=\:\mathrm{tan}\:\left(\pi−{C}\right) \\ $$$$\mathrm{tan}\:\left({A}+{B}\right)\:=\:−\mathrm{tan}\:{C} \\ $$

Commented by john santu last updated on 03/Jan/20

(2) tan (A+B) + tan C = 0  ((tan A+tan B)/(1−tan A×tan B))+tan C = 0  tan A+tan B+tan C−tan A×tan B×tan C = 0  tan A+tan B+tan C = tan A×tan B×tan C

$$\left(\mathrm{2}\right)\:\mathrm{tan}\:\left({A}+{B}\right)\:+\:\mathrm{tan}\:{C}\:=\:\mathrm{0} \\ $$$$\frac{\mathrm{tan}\:{A}+\mathrm{tan}\:{B}}{\mathrm{1}−\mathrm{tan}\:{A}×\mathrm{tan}\:{B}}+\mathrm{tan}\:{C}\:=\:\mathrm{0} \\ $$$$\mathrm{tan}\:{A}+\mathrm{tan}\:{B}+\mathrm{tan}\:{C}−\mathrm{tan}\:{A}×\mathrm{tan}\:{B}×\mathrm{tan}\:{C}\:=\:\mathrm{0} \\ $$$$\mathrm{tan}\:{A}+\mathrm{tan}\:{B}+\mathrm{tan}\:{C}\:=\:\mathrm{tan}\:{A}×\mathrm{tan}\:{B}×\mathrm{tan}\:{C} \\ $$

Commented by mathocean1 last updated on 03/Jan/20

thank you sir...

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}... \\ $$

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