If `((1+cosphi+isinphi)/(1+cosphi-isinphi))^n=u+i v` ,where `u and v` all real numbers ,then u is

If ((1+cos phi+i sin phi)/(1+cos phi v-i sin phi))^(n)=u+iv, where u and v all real number,then u is :

[frac{1+cosphi+isinphi}{1+cosphi-isinphi}]^n =

Show that ((1+cosphi+isinphi)/(1+cosphi-isinphi))^n=cosnphi+i sin nphi

Show that ((1+cosphi+isinphi)/(1+cosphi-isinphi))^n=cosphinphi+i sin nphi

If u=(cosx+sinx)/(cosx-sinx)=(1)/(v)," then "u'-v'=

If a = cos theta + isin theta , b = cosphi + isinphi , c = cos psi + i sin psi and a/b+b/c+c/a=2 then sin(theta-phi)+sin(phi-psi)+sin(psi-theta) equals

If a = cos theta + isin theta , b = cosphi + isinphi , c = cos psi + i sin psi and a/b+b/c+c/a=2 then sin(theta-phi)+sin(phi-psi)+sin(psi-theta) equals

If a = cos theta + isin theta , b = cosphi + isinphi , c = cos psi + i sin psi and a/b+b/c+c/a=2 then sin(theta-phi)+sin(phi-psi)+sin(psi-theta) equals

The focal f to a mirror is given by (1)/(f) = (1)/(u) + (1)/(v) where u and v represent object and image distance respectively then

If (x , y) and (x ,y) are the coordinates of the same point referred to two sets of rectangular axes with the same origin and it u x+v y , where u and v are independent of xa n dy , becomes V X+U Y , show that u^2+v^2=U^2+V^2dot