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

((costheta+isintheta)^5(cos3theta-isin3theta)^6)/((cos2theta+isin2theta)^3(cos4theta-isin4theta)^5)=

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

frac{(cos2theta-isin2theta)^4(cos4theta+isin4theta)^-5}{(cos3theta+isin3theta)^-2(cos3theta-isin3theta)^-9} =

Show that, ((cos2theta+isin2theta)^3(cos3theta-isin3theta)^4)/((cos3theta+isin3theta)^2(cos4theta+isin4theta)^(-3))=1

Simplify : ((cos3theta+isin3theta)^7(cos5theta-isin5theta)^4)/((cos4theta+isin4theta)^10(cos13theta-isin13theta)^3)

Simplify : ((cos3theta+isin3theta)^(2)(cos3theta-isin3theta)^(-3))/((cos4theta+isin4theta)^(-4)(costheta+isintheta)^(2))

If theta= (pi)/(6) , then the 10th term of the series 1+(cos theta +isin theta)+(cos theta +isin theta)^(2)+"……." is

Use De Moivre's theorem and simplify the following (cos2theta+isin2theta)^7/(cos4theta+isin4theta)^3

((costheta+isintheta)^(5)(cos3theta-isin3theta)^(6))/((cos2theta+isin2theta)^(3)(cos4theta-isin4theta)^(5))

The general value of theta which satisfies the equation (costheta+isintheta)(cos3theta+isin3theta) (cos5theta+isin5theta)…………((cos2n-1)theta+isin(2n-1)theta) = 1 is