Question

Put in the form A +iB

$\frac{(\cos x+i\sin x)(\cos y+i\sin y)}{[\cot u+i](1+i\tan v)}$

• A. $\sin u\cos v[\cos (x+y-u-v)-i\sin (x+y-u-v)]$
• B. $\sin u\cos v[\cos (x+y-u-v)+i\sin (x+y-u-v)]$
• C. $\sin v\cos u[\cos (x+y-u-v)+i\sin (x+y-u-v)]$
• D. $\sin v\cos u[\cos (x+y-u-v)-i\sin (x+y-u-v)]$

Answer 1

The correct option is B $\sin u\cos v[\cos (x+y-u-v)+i\sin (x+y-u-v)]$

Given expression

$=\frac{(\cos x+i\sin x)(\cos y+i\sin y)}{(\cos u/\sin u+i)(1+i\sin v/\cos v)}$

$=\frac{(\cos x+i\sin x)(\cos y+i\sin y)}{(\cos u+i\sin u)(\cos v+i\sin v)}\sin u\cos v$

$=\frac{\cos (x+y)+i\sin (x+y)}{\cos (u+v)+i\sin (u+v)}.\sin u\cos v$

$=\frac{e^{i(x+y)}}{e^{i(u+v)}}\sin u\sin v=e^{i(x+y-u-v)}\times \sin u\cos v$

$=\sin u\cos v[\cos (x+y-u-v)+i\sin (x+y-u-v)]$
Ans: B