Question:
The value of $ \frac{\cos 30{}^\circ +i\sin 30{}^\circ }{\cos 60{}^\circ -i\sin 60{}^\circ } $ is equal to
The value of $ \frac{\cos 30{}^\circ +i\sin 30{}^\circ }{\cos 60{}^\circ -i\sin 60{}^\circ } $ is equal to
Updated On: Jun 6, 2022
- $ i $
- $ -i $
- $ \frac{1+\sqrt{3}i}{2} $
- $ \frac{1-\sqrt{3}i}{2} $
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The Correct Option is A
Solution and Explanation
$ LHS=\frac{\cos 30{}^\circ +i\sin 30{}^\circ }{\cos 60{}^\circ -i\sin 60{}^\circ } $
$=\frac{\frac{\sqrt{3}}{2}+i\frac{1}{2}}{\frac{1}{2}-i\frac{\sqrt{3}}{2}}=\frac{\frac{\sqrt{3}+i}{2}}{\frac{1-i\sqrt{3}}{2}} $
$=\frac{\sqrt{3}+i}{1-i\sqrt{3}}\times \frac{1+i\sqrt{3}}{1+i\sqrt{3}}=\frac{\sqrt{3}+3i+i-\sqrt{3}}{1+3} $
$=\frac{4i}{4}=i $
$=\frac{\frac{\sqrt{3}}{2}+i\frac{1}{2}}{\frac{1}{2}-i\frac{\sqrt{3}}{2}}=\frac{\frac{\sqrt{3}+i}{2}}{\frac{1-i\sqrt{3}}{2}} $
$=\frac{\sqrt{3}+i}{1-i\sqrt{3}}\times \frac{1+i\sqrt{3}}{1+i\sqrt{3}}=\frac{\sqrt{3}+3i+i-\sqrt{3}}{1+3} $
$=\frac{4i}{4}=i $
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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
