Question

If we have the expression as $y=\cos \theta +i\sin \theta $ then the value of $y+\dfrac{1}{y}$ is
a). $2\cos \theta $
b). $2\sin \theta $
c). $2\cos ec\theta $
d). $2\tan \theta $

Answer 1

Hint: So we need to find the value of $y+\dfrac{1}{y}$ in which the value of ‘y’ is given in the trigonometric form with an imaginary number. So we will try to eliminate the imaginary function from the denominator, by using the method of rationalisation, and then we will simply solve it using trigonometric identities.

Complete step-by-step solution:
Moving ahead with the question in the step wise manner, we have $y=\cos \theta +i\sin \theta $ and we need to find out the value of $y+\dfrac{1}{y}$. So let us first put the value of ‘y’ in $y+\dfrac{1}{y}$, so we will get;
$=\cos \theta +i\sin \theta +\dfrac{1}{\cos \theta +i\sin \theta }$
Since the value of ‘y’ is in the trigonometric function additionally with the imaginary function, since we cannot solve the imaginary number in the denominator, let us rationalise it to remove the imaginary part from the denominator.
So let us rationalise $\dfrac{1}{y}$ by simultaneously multiplying and dividing the term with $=\cos \theta -i\sin \theta $, so we will get;
$=\cos \theta +i\sin \theta +\dfrac{1}{\cos \theta +i\sin \theta }\left( \dfrac{\cos \theta -i\sin \theta }{\cos \theta -i\sin \theta } \right)$
On simplifying it further we will get;
$=\cos \theta +i\sin \theta +\dfrac{\cos \theta -i\sin \theta }{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }$
As by the trigonometric identity we know that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta $ is equal to 1, so by replacing ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta $ with 1 we will get;
$=\cos \theta +i\sin \theta +\dfrac{\cos \theta -i\sin \theta }{1}$
On simplifying we will get;
$\begin{align}
  & \Rightarrow \cos \theta +i\sin \theta +\cos \theta -i\sin \theta \\
 & \Rightarrow 2\cos \theta \\
\end{align}$
So we got $2\cos \theta $.
Hence the answer is $2\cos \theta $ i.e. option ‘a’ is correct.

Note: Whenever the imaginary number comes in the denominator then our first step should be to rationalise it. Rationalisation means we should multiply and divide the number by term such that the denominator value will get reduced in the simpler form.