UPSC NDA I 2016 Mathematics Solved Paper

Concept:In a triangle, $A + B + C = 180^\circ$, so each cosine term transforms to a sine of half-angle using the complementary angle identity.Explanation:We have $\cos\left(\frac{A+B}{2}\right) = \cos\left(90^\circ - \frac{C}{2}\right) = \sin\frac{C}{2}$.Similarly, $\cos\left(\frac{B+C}{2}\right) = \sin\frac{A}{2}$ and $\cos\left(\frac{C+A}{2}\right) = \sin\frac{B}{2}$.Thus the product becomes $\sin\frac{A}{2}\;\sin\frac{B}{2}\;\sin\frac{C}{2}$.From the previous part of the question (or a known result in the context), $\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} = \frac{1}{8}$.Hence the value of the given expression is $\frac{1}{8}$.Answer:Since $\frac{1}{8}$ is not among the given options, the correct choice is D (None of the above).