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Question: Express in the form \[A + iB\] \[\left( {\dfrac{{1 + \sin \alpha + i\cos \alpha }}{{1 + \sin \alph...

Express in the form A+iBA + iB
(1+sin⁡α+icos⁡α1+sin⁡α−icos⁡α)\left( {\dfrac{{1 + \sin \alpha + i\cos \alpha }}{{1 + \sin \alpha - i\cos \alpha }}} \right)
A) cos⁡(nπ2−nα)−isin⁡(nπ2−nα)\cos \left( {\dfrac{{n\pi }}{2} - n\alpha } \right) - i\sin \left( {\dfrac{{n\pi }}{2} - n\alpha } \right)
B) cos⁡(nπ−nα)−isin⁡(nπ−nα)\cos \left( {n\pi - n\alpha } \right) - i\sin \left( {n\pi - n\alpha } \right)
C) cos⁡(nπ2−nα)+isin⁡(nπ2−nα)\cos \left( {\dfrac{{n\pi }}{2} - n\alpha } \right) + i\sin \left( {\dfrac{{n\pi }}{2} - n\alpha } \right)

Explanation

Solution

We are required to express the given expression in the A+iBA + 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. sin⁡θ=cos⁡(90−θ) =cos⁡(π2−θ)\begin{array}{l}\sin \theta = \cos \left( {90 - \theta } \right)\\\ = \cos \left( {\dfrac{\pi }{2} - \theta } \right)\end{array}
  2. cos⁡θ=sin⁡(90−θ) =sin⁡(π2−θ)\begin{array}{l}\cos \theta = \sin \left( {90 - \theta } \right)\\\ = \sin \left( {\dfrac{\pi }{2} - \theta } \right)\end{array}
  3. 1+cos⁡2θ=2cos⁡2θ1 + \cos 2\theta = 2{\cos ^2}\theta
  4. sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin \theta \cos \theta
  5. eiθ=cos⁡θ+isin⁡θ{e^{i\theta }} = \cos \theta + i\sin \theta
  6. e−iθ=cos⁡θ−isin⁡θ{e^{ - i\theta }} = \cos \theta - i\sin \theta

Complete step by step solution:
Let us substitute sin⁡α\sin \alpha and cos⁡α\cos \alpha in the given expression using the identities sin⁡θ=cos⁡(π2−θ)\sin \theta = \cos \left( {\dfrac{\pi }{2} - \theta } \right) and cos⁡θ=sin⁡(π2−θ)\cos \theta = \sin \left( {\dfrac{\pi }{2} - \theta } \right) respectively.
Consider sin⁡θ=cos⁡(π2−θ)\sin \theta = \cos \left( {\dfrac{\pi }{2} - \theta } \right). Substitute θ\theta with α\alpha in this equation. On doing so we get,
sin⁡α=cos⁡(π2−α)\sin \alpha = \cos \left( {\dfrac{\pi }{2} - \alpha } \right)
Consider cos⁡θ=sin⁡(π2−θ)\cos \theta = \sin \left( {\dfrac{\pi }{2} - \theta } \right). Substitute θ\theta with α\alpha in this equation. On doing so we get,
cos⁡α=sin⁡(π2−α)\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\\}……………….(1)\left( 1 \right)
We will now use the identity 1+cos⁡2θ=2cos⁡2θ1 + \cos 2\theta = 2{\cos ^2}\theta to substitute 1+cos⁡(π2−α)1 + \cos \left( {\dfrac{\pi }{2} - \alpha } \right) in the above expression.
Let us compare 1+cos⁡2θ1 + \cos 2\theta with 1+cos⁡(π2−α)1 + \cos \left( {\dfrac{\pi }{2} - \alpha } \right).
1+cos⁡2θ=1+cos⁡(π2−α) ⇒cos⁡2θ=cos⁡(π2−α) ⇒2θ=(π2−α)\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 2θ=(π2−α)2\theta = \left( {\dfrac{\pi }{2} - \alpha } \right). We will divide both sides of this equation with 2 to find the value of θ\theta . On doing so we get,
2θ2=12(π2−α) ⇒θ=(π4−α2)\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 2cos⁡2θ2{\cos ^2}\theta . To do so we will substitute θ=(π4−α2)\theta = \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right) in the expression.
2cos⁡2θ=2cos⁡2(π4−α2)2{\cos ^2}\theta = 2{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right).
Finally, we will substitute 2cos⁡2(π4−α2)2{\cos ^2}\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)with 1+cos⁡(π2−α)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\\}……………(2)\left( 2 \right)
We will now use the identity sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin \theta \cos \theta to substitute sin⁡(π2−α)\sin \left( {\dfrac{\pi }{2} - \alpha } \right) in the above expression. Let us compare sin⁡2θ\sin 2\theta with sin⁡(π2−α)\sin \left( {\dfrac{\pi }{2} - \alpha } \right).
sin⁡2θ=sin⁡(π2−α) ⇒2θ=(π2−α)\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 2θ=(π2−α)2\theta = \left( {\dfrac{\pi }{2} - \alpha } \right). We will divide both sides of this equation with 2 to find the value of θ\theta . On doing so we get,
2θ2=12(π2−α) ⇒θ=(π4−α2)\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 2sin⁡θcos⁡θ2\sin \theta \cos \theta . To do so we will substitute θ=(π4−α2)\theta = \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right) in the expression.
2sin⁡θcos⁡θ=2sin⁡(π4−α2)cos⁡(π4−α2)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 2sin⁡(π4−α2)cos⁡(π4−α2)2\sin \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)\cos \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)with sin⁡(π2−α)\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 θ=(π4−α2)\theta = \left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)
Now we also know that eiθ=cos⁡θ+isin⁡θ{e^{i\theta }} = \cos \theta + i\sin \theta and e−iθ=cos⁡θ−isin⁡θ{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,
ei(π4−α2)e−i(π4−α2)=ei(π4−α2)+i(π4−α2)\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
ei(π4−α2)+i(π4−α2)=e2i(π4−α2){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,
e2i(π4−α2)=ei(π2−α){e^{2i\left( {\dfrac{\pi }{4} - \dfrac{\alpha }{2}} \right)}} = {e^{i\left( {\dfrac{\pi }{2} - \alpha } \right)}}
On multiplying the power with nn, we get
en×i(π2−α)=ei(nπ2−nα){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 eiθ=cos⁡θ+isin⁡θ{e^{i\theta }} = \cos \theta + i\sin \theta . On doing so we will get,
ei(nπ2−nα)=cos⁡(nπ2−nα)+isin⁡(nπ2−nα){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+iBA + iB is in the form of a complex number.
Here, AA and BB are the real numbers. They are known as the real part.
ii denotes an imaginary number which is known as the imaginary part. The imaginary part satisfies the equation i2=−1{i^2} = - 1. No real number is able to satisfy this equation.