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# Express the following in the form A + iB .$\dfrac{1}{{1 - \cos \theta + 2i\sin \theta }}$

Last updated date: 17th Sep 2024
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Hint: Start by rationalizing the term . Use trigonometric identities and conversion formulas to simplify the fraction. Separate the real and imaginary part after simplification , the result obtained will be in the form of A + iB.

Given,
$\dfrac{1}{{1 - \cos \theta + 2i\sin \theta }}$
Let us start by rationalizing this equation, that is multiplying $1 - \cos \theta - 2i\sin \theta$ with the numerator and the denominator. We get
$\dfrac{{1 \times \left( {1 - \cos \theta - 2i\sin \theta } \right)}}{{\left( {1 - \cos \theta + 2i\sin \theta } \right) \times \left( {1 - \cos \theta - 2i\sin \theta } \right)}}$
As we know $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$, applying this formula in denominator , we get
$= \dfrac{{1 - \cos \theta - 2i\sin \theta }}{{{{\left( {1 - \cos \theta } \right)}^2} - {{\left( {2i} \right)}^2}\left( {{{\sin }^2}\theta } \right)}}$
We know ${i^2} = - 1\& {\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$, Applying this formula in denominator , we get
$\Rightarrow \dfrac{{1 - \cos \theta - 2i\sin \theta }}{{1 + {{\cos }^2}\theta - 2\cos \theta + 4{{\sin }^2}\theta }}$
We know that ,${\sin ^2}\theta + {\cos ^2}\theta = 1$
$\Rightarrow \dfrac{{1 - \cos \theta - 2i\sin \theta }}{{2 - 2\cos \theta + 3{{\sin }^2}\theta }}$
$\Rightarrow \dfrac{{1 - \cos \theta - 2i\sin \theta }}{{2(1 - \cos \theta ) + 3{{\sin }^2}\theta }}$
We know, $1 - \cos \theta = 2{\sin ^2}\dfrac{\theta }{2}$ and $\sin \theta = 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$
$\Rightarrow \dfrac{{2{{\sin }^2}\dfrac{\theta }{2} - i2 \cdot 2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}}{{2 \cdot 2{{\sin }^2}\dfrac{\theta }{2} + 3 \cdot {2^2}{{\sin }^2}\dfrac{\theta }{2}{{\cos }^2}\dfrac{\theta }{2}}}$
Taking $4{\sin ^2}\dfrac{\theta }{2}$in denominator and $4\sin \dfrac{\theta }{2}$in numerator as common , we get
$\Rightarrow \dfrac{{4\sin \dfrac{\theta }{2}(\sin \dfrac{\theta }{2} - i\cos \dfrac{\theta }{2})}}{{4{{\sin }^2}\dfrac{\theta }{2}(1 + 3{{\cos }^2}\dfrac{\theta }{2})}}$
On simplification , we get
$\Rightarrow \dfrac{{(\sin \dfrac{\theta }{2} - i\cos \dfrac{\theta }{2})}}{{\sin \dfrac{\theta }{2}(1 + 3{{\cos }^2}\dfrac{\theta }{2})}}$
Now separating the real and imaginary part , we get
$\dfrac{{\sin \dfrac{\theta }{2}}}{{\sin \dfrac{\theta }{2}(1 + 3{{\cos }^2}\dfrac{\theta }{2})}} - i\dfrac{{\cos \dfrac{\theta }{2}}}{{\sin \dfrac{\theta }{2}(1 + 3{{\cos }^2}\dfrac{\theta }{2})}}$
On further simplification ,we get
$\dfrac{1}{{(1 + 3{{\cos }^2}\dfrac{\theta }{2})}} - i\dfrac{{\cot \dfrac{\theta }{2}}}{{(1 + 3{{\cos }^2}\dfrac{\theta }{2})}}$
So this is in the form of A + iB , where A$= \dfrac{1}{{(1 + 3{{\cos }^2}\dfrac{\theta }{2})}}$and B$= - \dfrac{{\cot \dfrac{\theta }{2}}}{{(1 + 3{{\cos }^2}\dfrac{\theta }{2})}}$

Note: Similar questions can be solved by using the procedure , sometimes we just need to solve the exponent part first and sometimes we need to rationalize the terms. Students must know all the formulas related to trigonometric conversions. Attention must be given while substituting the values as it might lead to wrong answers.