Question

Solve the given complex function:
 $\dfrac{\left( \cos x+i\sin x \right)\left( \cos y+i\sin y \right)}{\left( \cot u+i \right)\left( 1+i\tan v \right)}$
A). $sinu\cos v$
B). $\sin u\cos v\left\{ \cos \left( x+y-u-v \right)+i\sin \left( x+y-u-v \right) \right\}$
C). $\sin u\cos v\left\{ \cos \left( x+y-u-v \right)-i\sin \left( x+y-u-v \right) \right\}$
D). None of these

Answer 1

Hint: In the given expression “i” is an iota in a complex number. Multiplication of numerator will yield$\cos \left( x+y \right)+i\sin \left( x+y \right)$and the denominator will yield$\dfrac{\cos \left( u+v \right)+i\sin \left( u+v \right)}{\sin u\cos v}$. Now, divide the numerator over the denominator and hence, we have got the answer.

Complete step-by-step solution -
The expression given in the question that we need to solve is:
 $\dfrac{\left( \cos x+i\sin x \right)\left( \cos y+i\sin y \right)}{\left( \cot u+i \right)\left( 1+i\tan v \right)}$
In the following, we are going to solve the numerator and denominator separately.
Solving numerator of the given expression we get,
$\begin{align}
  & \left( \cos x+i\sin x \right)\left( \cos y+i\sin y \right) \\
 & =\cos x\cos y+{{\left( i \right)}^{2}}\sin x\sin y+i\left( \cos x\sin y+\sin x\cos y \right) \\
\end{align}$
From the complex number, we know that${{i}^{2}}=-1$and then substituting in the above expression we get,
$\cos x\cos y-\sin x\sin y+i\left( \cos x\sin y+\sin x\cos y \right)$
From the trigonometric identities, we know that$\cos \left( x+y \right)=\cos x\cos y-\sin x\sin y$and$\sin \left( x+y \right)=\sin x\cos y+\cos x\sin y$. Substituting these values in the above expression we get,
$\cos \left( x+y \right)+i\sin \left( x+y \right)$
From the above simplification, the numerator is reduced to$\cos \left( x+y \right)+i\sin \left( x+y \right)$.
Now, solving the denominator of the expression given in the question we get,
$\begin{align}
  & \left( \cot u+i \right)\left( 1+i\tan v \right) \\
 & =\cot u+{{\left( i \right)}^{2}}\tan v+i\left( 1+\cot u\tan v \right) \\
\end{align}$
From the complex number, we know that${{i}^{2}}=-1$and then substituting in the above expression we get,
$\cot u-\tan v+i\left( 1+\cot u\tan v \right)$
In the above expression, we can write$\cot u=\dfrac{\cos u}{\sin u}$and$\tan v=\dfrac{\sin v}{\cos v}$then simplify the above expression.
$\begin{align}
  & \dfrac{\cos u}{\sin u}-\dfrac{\sin v}{\cos v}+i\left( 1+\dfrac{\cos u}{\sin u}\left( \dfrac{\sin v}{\cos v} \right) \right) \\
 & =\dfrac{\cos u\cos v-\sin u\sin v+i\left( \sin u\cos v+\cos u\sin v \right)}{\sin u\cos v} \\
\end{align}$
From the trigonometric identities, we know that$\cos \left( u+v \right)=\cos u\cos v-\sin u\sin v$and$\sin \left( u+v \right)=\sin u\cos v+\cos u\sin v$. Substituting these values in the above expression we get,
$\dfrac{\cos \left( u+v \right)+i\sin \left( u+v \right)}{\sin u\cos v}$
From the above simplification, the denominator is reduced to$\dfrac{\cos \left( u+v \right)+i\sin \left( u+v \right)}{\sin u\cos v}$.
Plugging these reduced values of numerator and denominator of the given expression we get,
$\begin{align}
  & \dfrac{\cos \left( x+y \right)+i\sin \left( x+y \right)}{\dfrac{\cos \left( u+v \right)+i\sin \left( u+v \right)}{\sin u\cos v}} \\
 & =\dfrac{\sin u\cos v\left( \cos \left( x+y \right)+i\sin \left( x+y \right) \right)}{\cos \left( u+v \right)+i\sin \left( u+v \right)} \\
\end{align}$
Now, multiplying and dividing the above expression by$\cos \left( u+v \right)-i\sin \left( u+v \right)$we get,
$\dfrac{\sin u\cos v\left( \cos \left( x+y \right)+i\sin \left( x+y \right) \right)}{\cos \left( u+v \right)+i\sin \left( u+v \right)}\times \dfrac{\cos \left( u+v \right)-i\sin \left( u+v \right)}{\cos \left( u+v \right)-i\sin \left( u+v \right)}$
$=\dfrac{\sin u\cos v\left( \cos \left( x+y \right)+i\sin \left( x+y \right) \right)\left( \cos \left( u+v \right)-i\sin \left( u+v \right) \right)}{{{\cos }^{2}}\left( u+v \right)+{{\sin }^{2}}\left( u+v \right)}$
From the trigonometric identities, we know that${{\cos }^{2}}\left( u+v \right)+{{\sin }^{2}}\left( u+v \right)=1$so using this relation in the above expression we get,
$\begin{align}
  & =\dfrac{\sin u\cos v\left( \cos \left( x+y \right)+i\sin \left( x+y \right) \right)\left( \cos \left( u+v \right)-i\sin \left( u+v \right) \right)}{1} \\
 & =\sin u\cos v\left( \cos \left( x+y \right)\cos \left( u+v \right)+\sin \left( x+y \right)\sin \left( u+v \right)+i\left( \sin \left( x+y \right)\cos \left( u+v \right)-\cos \left( x+y \right)\sin \left( u+v \right) \right) \right) \\
 & =\sin u\cos v\left\{ \cos \left( x+y-u-v \right)+i\sin \left( x+y-u-v \right) \right\} \\
\end{align}$From the above simplification, the given expression is resolved to:
$\sin u\cos v\left\{ \cos \left( x+y-u-v \right)+i\sin \left( x+y-u-v \right) \right\}$
Hence, the correct option is (b).

Note: The other way of solving the above problem is as follows:
First of all we are going to solve the denominator of the given expression in the form of$\cos \theta +i\sin \theta $:
$\begin{align}
  & \left( \cot u+i \right)\left( 1+i\tan v \right) \\
 & =\cot u+{{\left( i \right)}^{2}}\tan v+i\left( 1+\cot u\tan v \right) \\
\end{align}$
From the complex number, we know that${{i}^{2}}=-1$and then substituting in the above expression we get,
$\cot u-\tan v+i\left( 1+\cot u\tan v \right)$
In the above expression, we can write$\cot u=\dfrac{\cos u}{\sin u}$and$\tan v=\dfrac{\sin v}{\cos v}$then simplify the above expression.
$\begin{align}
  & \dfrac{\cos u}{\sin u}-\dfrac{\sin v}{\cos v}+i\left( 1+\dfrac{\cos u}{\sin u}\left( \dfrac{\sin v}{\cos v} \right) \right) \\
 & =\dfrac{\cos u\cos v-\sin u\sin v+i\left( \sin u\cos v+\cos u\sin v \right)}{\sin u\cos v} \\
\end{align}$
From the trigonometric identities, we know that$\cos \left( u+v \right)=\cos u\cos v-\sin u\sin v$and$\sin \left( u+v \right)=\sin u\cos v+\cos u\sin v$. Substituting these values in the above expression we get,
$\dfrac{\cos \left( u+v \right)+i\sin \left( u+v \right)}{\sin u\cos v}$
From the above simplification, the denominator is reduced to$\dfrac{\cos \left( u+v \right)+i\sin \left( u+v \right)}{\sin u\cos v}$.
Now, writing the above expression in place of the denominator in$\dfrac{\left( \cos x+i\sin x \right)\left( \cos y+i\sin y \right)}{\left( \cot u+i \right)\left( 1+i\tan v \right)}$.
$\begin{align}
  & \dfrac{\left( \cos x+i\sin x \right)\left( \cos y+i\sin y \right)}{\dfrac{\cos \left( u+v \right)+i\sin \left( u+v \right)}{\sin u\cos v}} \\
 & =\dfrac{\sin u\cos v\left( \cos x+i\sin x \right)\left( \cos y+i\sin y \right)}{\cos \left( u+v \right)+i\sin \left( u+v \right)} \\
\end{align}$
We know that the Euler form of the complex number is$z=\cos \theta +i\sin \theta $or $z={{e}^{i\theta }}$so we can write the above expression as:
$\begin{align}
  & \cos x+i\sin x={{e}^{ix}} \\
 & \cos y+i\sin y={{e}^{iy}} \\
 & \cos \left( u+v \right)+i\sin \left( u+v \right)={{e}^{i\left( u+v \right)}} \\
\end{align}$
Now, plugging the above values in $\dfrac{\sin u\cos v\left( \cos x+i\sin x \right)\left( \cos y+i\sin y \right)}{\cos \left( u+v \right)+i\sin \left( u+v \right)}$ we get,
$\begin{align}
  & \dfrac{\sin u\cos v{{e}^{ix}}\left( {{e}^{iy}} \right)}{{{e}^{i\left( u+v \right)}}} \\
 & =\dfrac{\sin u\cos v{{e}^{i\left( x+y \right)}}}{{{e}^{i\left( u+v \right)}}} \\
 & =\sin u\cos v {{e}^{i\left( x+y-u-v \right)}} \\
\end{align}$
In the above calculation, we have used the property that if base is same in multiplication then the powers are added like ${{e}^{ix}}\left( {{e}^{iy}} \right)={{e}^{i\left( x+y \right)}}$ and we have also used the property that if base is same in division then powers got subtracted like $\dfrac{{{e}^{i\left( x+y \right)}}}{{{e}^{i\left( u+v \right)}}}={{e}^{i\left( x+y-u-v \right)}}$.
Now, using Euler form we can write ${{e}^{i\left( x+y-u-v \right)}}=\cos \left( x+y-u-v \right)+i\sin \left( x+y-u-v \right)$in the above expression we get,
$\sin u\cos v\left\{ \cos \left( x+y-u-v \right)+i\sin \left( x+y-u-v \right) \right\}$
Hence, we have got the same answer as we have obtained above.