Question

$A+iB$ form of $\frac{(\cos x+i\sin x)(\cos y+i\sin y)}{(\cot u+i)(1+i\tan v)}$ is

• A. $\sin u\cos v\left[\cos \right(x+y-u-v)+i\sin (x+y-u-v\left)\right]$
• B. $\sin u\cos v\left[\cos \right(x+y+u+v)+i\sin (x+y+u+v\left)\right]$
• C. $\sin u\cos v\left[\cos \right(x+y+u+v)-i\sin (x+y+u+v\left)\right]$
• D. None of the above

Answer 1

Answer: (A)

$\sin u\cos v\left[\cos \right(x+y-u-v)+i\sin (x+y-u-v\left)\right]$

$\text{LHS}=\frac{(\cos x+i\sin x)(\cos y+i\sin y}{(\cos u+i\sin u)(\cos v+i\sin v)}\sin u\cos v$
$=\sin u\cos v{e}^{ix}\cdot e^{iy}\cdot e^{-iu}\cdot e^{-iv}$
$=\sin u\cos v\left[\cos \right(x+y-u-v]+i\sin (x+y-u-v)$.