Express $\dfrac{1}{\left( 1-\cos \theta +2i\sin \theta \right)}$ in the form $x+iy$.

Hint: The main key point in this problem is dividing the numerator and denominator by the conjugate of the complex number in the denominator. If $z=a+ib$ is a complex number, then the conjugate of this complex number is defined as \[\overline{z}=a-ib\]. Then by simplifying the obtained expression we can express it in the form $x+iy$.

Complete step by step answer:
Here, it is given,
$\Rightarrow \dfrac{1}{\left( 1-\cos \theta +2i\sin \theta \right)}$
Here, the complex number in the denominator is,
$\Rightarrow z=(1-\cos \theta )+i(2\sin \theta ).........(i)$.
The conjugate of $z$ is represented as,
$\Rightarrow \overline{z}=(1-\cos \theta )-i(2\sin \theta ).........(ii)$.
Now, we have to multiply the numerator and denominator of the expression $\dfrac{1}{\left( 1-\cos \theta +2i\sin \theta \right)}$by equation (ii).
Then we get,
$\Rightarrow \dfrac{1}{\left( 1-\cos \theta +2i\sin \theta \right)}\times \dfrac{\left( 1-\cos \theta -2i\sin \theta \right)}{\left( 1-\cos \theta -2i\sin \theta \right)}$
Simplifying the above equation, we get,
\[\begin{align}
  & \Rightarrow \dfrac{\left( 1-\cos \theta -2i\sin \theta \right)}{\left( {{\left( 1-\cos \theta \right)}^{2}}-{{\left( 2i\sin \theta \right)}^{2}} \right)} \\
 & \Rightarrow \dfrac{\left( 1-\cos \theta -2i\sin \theta \right)}{\left( 1+{{\cos }^{2}}\theta -2\cos \theta -\left( 4\times {{i}^{2}}\times {{\sin }^{2}}\theta \right) \right)}.........(iii) \\
\end{align}\]
In equation (iii) we have ${{i}^{2}}=-1$. Now, we can rewrite equation (iii) as,
\[\Rightarrow \dfrac{\left( 1-\cos \theta -2i\sin \theta \right)}{\left( 1+{{\cos }^{2}}\theta -2\cos \theta +4{{\sin }^{2}}\theta \right)}.........(iv)\]
Now, we know ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.Thus, we can rewrite the denominator of equation (iv) as,
\[\begin{align}
  & \Rightarrow \dfrac{\left( 1-\cos \theta -2i\sin \theta \right)}{\left( 1+{{\cos }^{2}}\theta +{{\sin }^{2}}\theta -2\cos \theta +3{{\sin }^{2}}\theta \right)} \\
 & \Rightarrow \dfrac{\left( 1-\cos \theta -2i\sin \theta \right)}{\left( 1+1-2\cos \theta +3{{\sin }^{2}}\theta \right)} \\
 & \Rightarrow \dfrac{\left( 1-\cos \theta -2i\sin \theta \right)}{\left( 2-2\cos \theta +3{{\sin }^{2}}\theta \right)}..........(v) \\
\end{align}\]
We also know that, ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $. Substituting this in equation (v) we get,
\[\Rightarrow \dfrac{\left( 1-\cos \theta -2i\sin \theta \right)}{\left( 2-2\cos \theta +3\left( 1-{{\cos }^{2}}\theta \right) \right)}..........(vi)\]
Now, simplifying the above equation we get,
\[\begin{align}
  & \Rightarrow \dfrac{\left( 1-\cos \theta -2i\sin \theta \right)}{\left( 2-2\cos \theta +3-3{{\cos }^{2}}\theta \right)} \\
 & \Rightarrow \dfrac{\left( 1-\cos \theta -2i\sin \theta \right)}{\left( 5-2\cos \theta -3{{\cos }^{2}}\theta \right)}.........(vii) \\
\end{align}\]
Now, \[\left( 5-2\cos \theta -3{{\cos }^{2}}\theta \right)\] can be simplified as,
$\begin{align}
  & \Rightarrow \left( -\left( 3{{\cos }^{2}}\theta +2\cos \theta -5 \right) \right) \\
 & \Rightarrow -\left( 3{{\cos }^{2}}\theta -3\cos \theta +5\cos \theta -5 \right) \\
 & \Rightarrow -\left( 3\cos \theta \left( \cos \theta -1 \right)+5\left( \cos \theta -1 \right) \right) \\
 & \Rightarrow -\left( 3\cos \theta +5 \right)\left( \cos \theta -1 \right) \\
 & \Rightarrow \left( 1-\cos \theta \right)\left( 3\cos \theta +5 \right).........(viii) \\
\end{align}$
Substituting equation (viii) in equation (vii), we get,
\[\Rightarrow \dfrac{\left( 1-\cos \theta -2i\sin \theta \right)}{\left( 1-\cos \theta \right)\left( 3\cos \theta +5 \right)}.........(ix)\]
On simplifying equation (ix) we get,
\[\begin{align}
  & \Rightarrow \dfrac{\left( 1-\cos \theta \right)}{\left( 1-\cos \theta \right)\left( 3\cos \theta +5 \right)}-i\dfrac{2\sin \theta }{\left( 1-\cos \theta \right)\left( 3\cos \theta +5 \right)} \\
 & \Rightarrow \dfrac{{\left( 1-\cos \theta \right)}}{{\left( 1-\cos \theta \right)}\left( 3\cos \theta +5 \right)}-i\dfrac{2\sin \theta }{\left( 1-\cos \theta \right)\left( 3\cos \theta +5 \right)} \\
 & \\
\end{align}\]
\[\Rightarrow \dfrac{1}{\left( 3\cos \theta +5 \right)}-i\dfrac{2\sin \theta }{\left( 1-\cos \theta \right)\left( 3\cos \theta +5 \right)}............(x)\]
We know that,
$\begin{align}
  & \Rightarrow \sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}........(xi) \\
 & \Rightarrow 1-\cos \theta =2{{\sin }^{2}}\dfrac{\theta }{2}........(xii) \\
\end{align}$
Substituting equations (xi) and (xii) in equation (x) we get,
\[\Rightarrow \dfrac{1}{\left( 3\cos \theta +5 \right)}-i\dfrac{4\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}{2{{\sin }^{2}}\dfrac{\theta }{2}\left( 3\cos \theta +5 \right)}\]
On simplifying the above equation, we get,
\[\begin{align}
  & \Rightarrow \dfrac{1}{\left( 3\cos \theta +5 \right)}-i\dfrac{{4\sin \dfrac{\theta }{2}}\cos \dfrac{\theta }{2}}{{2}{{\sin }^{{2}}}\dfrac{\theta }{2}\left( 3\cos \theta +5 \right)} \\
 & \Rightarrow \dfrac{1}{\left( 3\cos \theta +5 \right)}-i\dfrac{2\cos \dfrac{\theta }{2}}{\sin \dfrac{\theta }{2}\left( 3\cos \theta +5 \right)} \\
 & \Rightarrow \dfrac{1}{\left( 3\cos \theta +5 \right)}-i\dfrac{2\cot \dfrac{\theta }{2}}{\left( 3\cos \theta +5 \right)}...........(xiii) \\
\end{align}\]
Now we have obtained the expression in the form of $x+iy$.
From equation (xiii) we know that,
The real part $x=\dfrac{1}{3\cos \theta +5}$ and the imaginary part is $-\dfrac{2\cot \dfrac{\theta }{2}}{3\cos \theta +5}$.
Hence, the correct answer is \[\dfrac{1}{\left( 3\cos \theta +5 \right)}-i\dfrac{2\cot \dfrac{\theta }{2}}{\left( 3\cos \theta +5 \right)}\].
Note: The student should know how to take the conjugate of a complex number and should also know the basic trigonometric functions in order to solve this problem. A complex number is equal to its complex conjugate if its imaginary part is zero.