`If a=cisalpha,b=cisbeta,c=cisgamma` then `(a^3b^3)/c^2=`

If a=cisalpha,b=cisbeta,c=cisgamma , and a/b+b/c+c/a-1=0 then cos(alpha-beta)+cos(beta-gamma) + cos(gamma-alpha) =

If x=cisalpha,y=cisbeta" then " x^3y^4-(1)/(x^3y^4)=

Match the following. {:(Column-I," "Column-II),(I" : "x=costheta+isintheta" then "x^n-(1)/(x^n),"a) "2isin(alpha-beta)),(II" : If " z=costheta+sintheta" then "(z^n-1)/(z^(2n)+1),"b) "2cos(alpha-beta)),(III" : "x=cisalpha","y=cisbeta" then "x/y+y/x=,"c) "itanntheta),(IV " : " x=cisalpha","y=cisbeta" then "x/y-y/x=,"d) "2iSinntheta):}

If a=-1,b=2,c=3," then"(a^(3)+b^(3)+c^(3)-3abc)/((a-b)^(2)+(b-c)^(2)+(c-a)^(2))=

If a= 25, b= 15, c= - 10 then (a^(3) + b^(3) + c^(3) - 3abc)/((a-b)^(2) + (b-c)^(2) + (c-a)^(2))

A : If A+B+C=piandcisA=x,cisB=y,cisC=z , then xyz=-1 R: (cisalpha)(cisbeta)=cis(alpha+beta)

If sides of triangle ABC are a,b and c such that 2b=a+c then (b)/(c)>(2)/(3)( b) (b)/(c)>(1)/(3)(b)/(c)<2 (d) (b)/(c)<(3)/(2)

Prove: |[b+c ,a-b ,a], [c+a, b-c ,b], [a+b, c-a, c]|=3a b c-a^3-b^3-c^3

If a , b ,c in Ra n da(a+b)+b(b+c)+c(c+a)=0 then a.a=b=c=0 b. a+b+c+=0 c. (a-b)^2+(b-c)^2+(c-a)^2=0 d. a^3+b^3+c^3+3a b c=0

If a, b, c are in A.P., prove that a^(3)+4b^(3)+c^(3)=3b(a^(2)+c^(2)).