Question

Express in the form \[A + iB\]
\[\left( {\dfrac{{1 + \sin \alpha + i\cos \alpha }}{{1 + \sin \alpha - i\cos \alpha }}} \right)\]
A) \[\cos \left( {\dfrac{{n\pi }}{2} - n\alpha } \right) - i\sin \left( {\dfrac{{n\pi }}{2} - n\alpha } \right)\]
B) \[\cos \left( {n\pi - n\alpha } \right) - i\sin \left( {n\pi - n\alpha } \right)\]
C) \[\cos \left( {\dfrac{{n\pi }}{2} - n\alpha } \right) + i\sin \left( {\dfrac{{n\pi }}{2} - n\alpha } \right)\]

Answer 1

Hint:
We are required to express the given expression in the \[A + iB\] form. To solve this question, we will first manipulate our expression using the various trigonometric identities and formulas to simplify it to the simplest form. We will then express it as an exponential and then simplify further to obtain the answer.
Formula Used: We will use the following formulas to solve our question,
1) \[\begin{array}{l}\sin \theta = \cos \left( {90 - \theta } \right)\\ = \cos \left( {\dfrac{\pi }{2} - \theta } \right)\end{array}\]
2) \[\begin{array}{l}\cos \theta = \sin \left( {90 - \theta } \right)\\ = \sin \left( {\dfrac{\pi }{2} - \theta } \right)\end{array}\]
3) \[1 + \cos 2\theta = 2{\cos ^2}\theta \]
4) \[\sin 2\theta = 2\sin \theta \cos \theta \]
5) \[{e^{i\theta }} = \cos \theta + i\sin \theta \]
6) \[{e^{ - i\theta }} = \cos \theta - i\sin \theta \]

Complete step by step solution:
Let us substitute \[\sin \alpha \]and \[\cos \alpha \] in the given expression using the identities \[\sin \theta = \cos \left( {\dfrac{\pi }{2} - \theta } \right)\] and \[\cos \theta = \sin \left( {\dfrac{\pi }{2} - \theta } \right)\] respectively.
Consider \[\sin \theta = \cos \left( {\dfrac{\pi }{2} - \theta } \right)\]. Substitute \[\theta \] with \[\alpha \]in this equation. On doing so we get,
\[\sin \alpha = \cos \left( {\dfrac{\pi }{2} - \alpha } \right)\]
Consider \[\cos \theta = \sin \left( {\dfrac{\pi }{2} - \theta } \right)\]. Substitute \[\theta \] with \[\alpha \]in this equation. On doing so we get,
\[\cos \alpha = \sin \left( {\dfrac{\pi }{2} - \alpha } \right)\]
So, now based on the above modifications, we will substitute the values in our expression. On doing so we get,
\[\left( {\dfrac{{1 + \sin \alpha + i\cos \alpha }}{{1 + \sin \alpha - i\cos \alpha }}} \right) = \left\{ {\dfrac{{1 + \cos \left( {\dfrac{\pi }{2} - \alpha } \right) + i\sin \left( {\dfrac{\pi }{2} - \alpha } \right)}}{{1 + \cos \left( {\dfrac{\pi }{2} - \alpha } \right) - i\sin \left( {\dfrac{\pi }{2} - \alpha } \right)}}} \right\}\]……………….\[\left( 1 \right)\]
We will now use the identity \[1 + \cos 2\theta = 2{\cos ^2}\theta \] to substitute \[1 + \cos \left( {\dfrac{\pi }{2} - \alpha } \right)\] in the above expression.
Let us compare \[1 + \cos 2\theta \] with \[1 + \cos \left( {\dfrac{\pi }{2} - \alpha } \right)\].
\[\begin{array}{l}1 + \cos 2\theta = 1 + \cos \left( {\dfrac{\pi }{2} - \alpha } \right)\\ \Rightarrow \cos 2\theta = \cos \left( {\dfrac{\pi }{2} - \alpha } \right)\\ \Rightarrow 2\theta = \left( {\dfrac{\pi }{2} - \alpha } \right)\end{array}\]
Hence, we see that \[2\theta = \left( {\dfrac{\pi }{2} - \alpha } \right)\]. We will divide both sides of this equation with 2 to find the value of \[\theta \]. On doing so we get,
\[\begin{array}{l}\dfrac{{2\theta }}{2} = \dfrac{1}{2}\left( {\dfrac{\pi }{2} - \alpha } \right)\\ \Rightarrow \theta = \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)\end{array}\]
We will now find the value of \[2{\cos ^2}\theta \]. To do so we will substitute \[\theta = \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)\] in the expression.
\[2{\cos ^2}\theta = 2{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)\].
Finally, we will substitute \[2{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)\]with \[1 + \cos \left( {\dfrac{\pi }{2} - \alpha } \right)\] in equation (1). Hence, we get,
\[\left\{ {\dfrac{{1 + \cos \left( {\dfrac{\pi }{2} - \alpha } \right) + i\sin \left( {\dfrac{\pi }{2} - \alpha } \right)}}{{1 + \cos \left( {\dfrac{\pi }{2} - \alpha } \right) - i\sin \left( {\dfrac{\pi }{2} - \alpha } \right)}}} \right\} = \left\{ {\dfrac{{2{{\cos }^2}\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right) + i\sin \left( {\dfrac{\pi }{2} - \alpha } \right)}}{{2{{\cos }^2}\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right) - i\sin \left( {\dfrac{\pi }{2} - \alpha } \right)}}} \right\}\]……………\[\left( 2 \right)\]
We will now use the identity \[\sin 2\theta = 2\sin \theta \cos \theta \] to substitute \[\sin \left( {\dfrac{\pi }{2} - \alpha } \right)\] in the above expression. Let us compare \[\sin 2\theta \] with \[\sin \left( {\dfrac{\pi }{2} - \alpha } \right)\].
\[\begin{array}{l}\sin 2\theta = \sin \left( {\dfrac{\pi }{2} - \alpha } \right)\\ \Rightarrow 2\theta = \left( {\dfrac{\pi }{2} - \alpha } \right)\end{array}\]
Hence, we see that \[2\theta = \left( {\dfrac{\pi }{2} - \alpha } \right)\]. We will divide both sides of this equation with 2 to find the value of \[\theta \]. On doing so we get,
\[\begin{array}{l}\dfrac{{2\theta }}{2} = \dfrac{1}{2}\left( {\dfrac{\pi }{2} - \alpha } \right)\\ \Rightarrow \theta = \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)\end{array}\]
We will now find the value of \[2\sin \theta \cos \theta \]. To do so we will substitute \[\theta = \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)\] in the expression.
\[2\sin \theta \cos \theta = 2\sin \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)\cos \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)\].
Finally, we will substitute \[2\sin \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)\cos \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)\]with \[\sin \left( {\dfrac{\pi }{2} - \alpha } \right)\] in equation (2). Hence, we get,
\[\left\{ {\dfrac{{2{{\cos }^2}\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right) + i\sin \left( {\dfrac{\pi }{2} - \alpha } \right)}}{{2{{\cos }^2}\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right) - i\sin \left( {\dfrac{\pi }{2} - \alpha } \right)}}} \right\} = \left\{ {\dfrac{{2{{\cos }^2}\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right) + i2\sin \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)\cos \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)}}{{2{{\cos }^2}\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right) - i2\sin \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)\cos \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)}}} \right\}\]
We will now take the common terms out from the numerator and denominator of the above expression. On doing so we get,
\[\left\{ {\dfrac{{2{{\cos }^2}\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right) + i2\sin \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)\cos \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)}}{{2{{\cos }^2}\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right) - i2\sin \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)\cos \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)}}} \right\} = \dfrac{{2\cos \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)\left\{ {2\cos \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right) + i\sin \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)} \right\}}}{{2\cos \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)\left\{ {2\cos \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right) - i\sin \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)} \right\}}}\]
On canceling out the common terms we obtain the expression as,
\[\dfrac{{2\cos \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)\left\{ {\cos \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right) + i\sin \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)} \right\}}}{{2\cos \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)\left\{ {\cos \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right) - i\sin \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)} \right\}}} = \dfrac{{\left\{ {\cos \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right) + i\sin \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)} \right\}}}{{\left\{ {\cos \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right) - i\sin \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)} \right\}}}\]
We know that \[\theta = \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)\]
Now we also know that \[{e^{i\theta }} = \cos \theta + i\sin \theta \] and \[{e^{ - i\theta }} = \cos \theta - i\sin \theta \]. Thus, expressing the above equation, in the exponential form we will get,
\[\dfrac{{\left\{ {\cos \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right) + i\sin \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)} \right\}}}{{\left\{ {\cos \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right) - i\sin \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)} \right\}}} = \dfrac{{{e^{i\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)}}}}{{{e^{ - i\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)}}}}\]
On evaluating the powers of the above exponent, we get,
\[\dfrac{{{e^{i\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)}}}}{{{e^{ - i\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)}}}} = {e^{i\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right) + i\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)}}\]
On adding the powers, we get
\[{e^{i\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right) + i\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)}} = {e^{2i\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)}}\]
On multiplying the terms and simplifying the powers we get,
\[{e^{2i\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)}} = {e^{i\left( {\dfrac{\pi }{2} - \alpha } \right)}}\]
On multiplying the power with \[n\], we get
\[{e^{n \times i\left( {\dfrac{\pi }{2} - \alpha } \right)}} = {e^{i\left( {\dfrac{{n\pi }}{2} - n\alpha } \right)}}\]
Finally, we will express our exponential into the complex equation form using the formula \[{e^{i\theta }} = \cos \theta + i\sin \theta \]. On doing so we will get,
\[{e^{i\left( {\dfrac{{n\pi }}{2} - n\alpha } \right)}} = \cos \left( {\dfrac{{n\pi }}{2} - n\alpha } \right) + i\sin \left( {\dfrac{{n\pi }}{2} - n\alpha } \right)\]

Hence, the correct answer is option (C).

Note:
\[A + iB\] is in the form of a complex number.
Here, \[A\] and \[B\] are the real numbers. They are known as the real part.
\[i\] denotes an imaginary number which is known as the imaginary part. The imaginary part satisfies the equation \[{i^2} = - 1\]. No real number is able to satisfy this equation.