In the form of `A+iB` `z=(cos2theta+isin2theta)^(-5)(cos3theta-i sin3theta)^6(sintheta-icostheta)^3` is

(cos2theta+isin2theta)^(-5)(cos3theta-isin3theta)^(6)(sintheta-icostheta)^(3) is

((cos3theta-isin3theta)^4)/((sin2theta+icos2theta)^5)=

(cos3theta+isin3theta)^5/(costheta+isintheta)^6

(sin3theta-cos3theta)/(sintheta+costheta)+1 =

(3costheta+cos3theta)/(3sintheta-sin3theta)=

Statement-I : ((costheta+isintheta)^5(cos3theta-isin3theta)^6)/((cos2theta+isin2theta)^3(cos4theta-isin4theta)^5)=1 Statement-II : ((cos2theta+isin2theta)^3(cos3theta-isin3theta)^4)/((cos3theta+isin3theta)^2(cos4theta+isin4theta)^(-3))=1

Express the following in a+ib form: ((cos2 theta-i sin2 theta)^(4)(cos4 theta+i sin4 theta)^(-5))/((cos3 theta+i sin3 theta)^(-2)(cos3 theta-i sin3 theta)^(-9))

(cos^(3)theta-cos3theta)/(costheta)+(sin^(3)theta+sin3theta)/(sintheta)=

(sintheta-2sin^3theta)/(2cos^3theta-costheta) = …………….