UPSC NDA 2 2022 Mathematics Solved Paper

Concept:A complex number $z = x + iy$ is purely real if its imaginary part is zero, i.e., $y = 0$.Explanation:Step 1: Start with $z = \frac{1 + i\sin\theta}{1 - i\sin\theta}$.Step 2: Multiply numerator and denominator by $(1 + i\sin\theta)$ to rationalize:$z = \frac{(1 + i\sin\theta)^2}{(1 - i\sin\theta)(1 + i\sin\theta)}$.Step 3: Simplify using $(a+b)^2$ and $(a-b)(a+b) = a^2 - b^2$:Numerator becomes $1 + 2i\sin\theta - \sin^2\theta$.Denominator becomes $1 - (i\sin\theta)^2 = 1 + \sin^2\theta$.Step 4: Separate into real and imaginary parts:$z = \frac{1 - \sin^2\theta}{1 + \sin^2\theta} + i \cdot \frac{2\sin\theta}{1 + \sin^2\theta}$.Step 5: For $z$ to be purely real, the imaginary part must be zero:$\frac{2\sin\theta}{1 + \sin^2\theta} = 0$.Step 6: The denominator is always positive, so $\sin\theta = 0$.Step 7: The general solution is $\theta = n\pi$, where $n$ is an integer.Answer:$\theta = n\pi$, which corresponds to option C.