Prove by mathematical induction `(costheta+isintheta)^(n)=cosntheta+isinntheta` where `i=sqrt(-1)`.

Prove by mathematical induction n^(2) + n is even.

choose the correct pair of statement : (i) (costheta+isintheta)^(n)=cosntheta+isinntheta , x is an integer (ii) (sintheta+icostheta)^(n)=sinntheta+icosthetantheta , n is an integer (iii) If z=costheta+isintheta , then z^(-1)=costheta-isintheta (iv)(z+barz)/(2)=Re(z)

By mathematical induction prove that 2^(3n) -1 is divisible by 7.

If A=((costheta, sintheta), (-sintheta, costheta)) prove that AA^(T)=I

Prove by mathematical induction 4 + 8 + 12 +... 4n = 2n(n + 1).

Prove that identity sqrt((1-costheta)/(1+costheta))=cosectheta-cottheta

By mathematical induction show that 7^(2n)+16n-1 is divisible by 64.

Solve sintheta + sqrt3 costheta =1

Using the mathematical induction, show that for any natural number n, with the assumption i^(2)=-1 , (r(costheta+isintheta))^(n)=r^(n)(cosntheta+isinntheta)

If (costheta +isintheta)(cos2theta +isin2theta).....(cosntheta + isinntheta) = 1 then the value of theta is: