UPSC NDA I 2015 Mathematics Paper

Concept:The problem uses Euler's formula $e^{ix} = \cos x + i \sin x$ and trigonometric identities to simplify a complex expression raised to a power.Explanation:We start with the given expression: $\left[ \frac{\sin\frac{\pi}{6} + i(1-\cos\frac{\pi}{6})}{\sin\frac{\pi}{6} - i(1-\cos\frac{\pi}{6})} \right]^{3}$.Use the identities: $1 - \cos\theta = 2\sin^{2}(\theta/2)$ and $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$.Substitute $\theta = \pi/6$, so $\theta/2 = \pi/12$. The numerator becomes: $2\sin(\pi/12)\cos(\pi/12) + i(2\sin^{2}(\pi/12))$.Factor $2\sin(\pi/12)$: numerator $= 2\sin(\pi/12)[\cos(\pi/12) + i\sin(\pi/12)]$.Similarly, denominator: $2\sin(\pi/12)\cos(\pi/12) - i(2\sin^{2}(\pi/12)) = 2\sin(\pi/12)[\cos(\pi/12) - i\sin(\pi/12)]$.Cancel $2\sin(\pi/12)$ (non-zero). The expression becomes: $\left[ \frac{\cos(\pi/12) + i\sin(\pi/12)}{\cos(\pi/12) - i\sin(\pi/12)} \right]^{3}$.Apply Euler's formula: $\cos x + i\sin x = e^{ix}$, and $\cos x - i\sin x = e^{-ix}$. So we have $\left[ \frac{e^{i\pi/12}}{e^{-i\pi/12}} \right]^{3}$.Simplify inside: $\frac{e^{i\pi/12}}{e^{-i\pi/12}} = e^{i\pi/12} \cdot e^{i\pi/12} = e^{i\pi/6}$.Raise to the third power: $(e^{i\pi/6})^{3} = e^{i\pi/2}$.Finally, $e^{i\pi/2} = \cos(\pi/2) + i\sin(\pi/2) = 0 + i(1) = i$.Thus the value is $i$.Answer:$i$ (Option B).