Question

a+ib form of (cosx+isinx)(cosy+isiny)/(cotu+i)(1+itanv)​

Answer 1

Answer:

$\frac{(cosx+i\:sinx)(cosy+i\:siny)}{(cotu+i)(1+i\:tanv)}=(sinu\:cosv)cos((x+y)-(u+v))+i\:(sinu\:cosv)sin((x+y)-(u+v))$

Step-by-step explanation:

Euler's formula:

$\boxed{e^{i\theta}=cos\theta+i\:sin\theta}$

Given:

$\frac{(cosx+i\:sinx)(cosy+i\:siny)}{(cotu+i)(1+i\:tanv)}$

$=\frac{e^{ix}.e^{iy}}{(\frac{cosu}{sinu}+i)(1+i\frac{cosv}{sinv}})$

$=\frac{e^{i(x+y)}}{(\frac{cosu+i\:sinu}{sinu})(\frac{cosv+i\:sinv}{cosv})}$

$=\frac{e^{i(x+y)}}{\frac{e^{iu}}{sinu}.\frac{e^{iv}}{cosv}}$

$=\frac{e^{i(x+y)}}{\frac{e^{i(u+V)}}{sinu\:cosv}}$

$=\frac{e^{i(x+y)}.e^{-i(u+v)}}{sinu\:cosv}$

$=e^{i((x+y)-(u+v))}(sinu\:cosv)$

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

$=(sinu\:cosv)cos((x+y)-(u+v))+i\:(sinu\:cosv)sin((x+y)-(u+v))$

$\implies\:\frac{(cosx+i\:sinx)(cosy+i\:siny)}{(cotu+i)(1+i\:tanv)}=(sinu\:cosv)cos((x+y)-(u+v))+i\:(sinu\:cosv)sin((x+y)-(u+v))$