Suppose A, B, C are angles of a triangle, and let Δ$$=e2iAe$$−iCe−iBe−$$iCe2iBe$$−iAe−iBe−$$iAe2iC$$ then value of ∆ is

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Suppose A, B, C are angles of a triangle, and let Δ$$=e2iAe$$−iCe−iBe−$$iCe2iBe$$−iAe−iBe−$$iAe2iC$$ then value of ∆ is

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Talking eiA, common from $$R1$$,eiBfrom $$R2$$ and eiCfrom $$R3$$ we getΔ$$=ei(A+B+C)$$Δ$$1where$$ Δ$$1=eiAei(A+C)ei(A+B)e$$−$$i(B+C)eiBe$$−$$i(A+B)e$$−$$i(B+C)e$$−$$i(A+C)eiCBut$$ A $$+$$ B $$+$$ C $$=$$π, so that $$ei(A+R+C)=ei$$π$$=$$$$cos$$⁡π$$+isin$$⁡π$$=$$−$$1$$.$$AlsoA+C=$$π−B ⇒ e−$$i(A+C)=e$$−π$$ieiB=$$−eiBThus, Δ$$1=eiA$$−eiB−eiC−e−iAeiB−eiC−eiA−$$eiBeiC=ei(A+B+C)1$$−$$1$$−$$1$$−$$11$$−$$1$$−$$1$$−$$11Using$$ $$C1$$→$$C1+C2$$ we getΔ$$1=($$−$$1)0$$−$$1$$−$$101$$−$$1$$−$$2$$−$$11=($$−$$1)($$−$$2)(2)=4Therefor$$ Δ$$=($$−$$1)$$Δ$$1=$$−$$4$$

Suppose $A, B, C$ are angles of a triangle, and let $Δ = |\begin{array}{ccc} e^{2 i A} & e^{- i C} & e^{- i B} \\ e^{- i C} & e^{2 i B} & e^{- i A} \\ e^{- i B} & e^{- i A} & e^{2 i C} \end{array}|$ then value of $∆$ is

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Suppose A, B, C are angles of a triangle, and let Δ=e2iAe−iCe−iBe−iCe2iBe−iAe−iBe−iAe2iC then value of ∆ is

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Suppose $A, B, C$ are angles of a triangle, and let $Δ = |\begin{array}{ccc} e^{2 i A} & e^{- i C} & e^{- i B} \\ e^{- i C} & e^{2 i B} & e^{- i A} \\ e^{- i B} & e^{- i A} & e^{2 i C} \end{array}|$ then value of $∆$ is