UPSC NDA I 2019 Mathematics Solved Paper

Concept:Use trigonometric transformations for angles greater than $90^\circ$ and the identities $\sin(A+B)$ and $\cos(A-B)$ to simplify the given expression.Explanation:First, simplify each term using angle reduction formulas.For the numerator: $\cos 236^\circ = \cos(180^\circ + 56^\circ) = -\cos 56^\circ$ and $\sin 124^\circ = \sin(90^\circ + 34^\circ) = \cos 34^\circ$.Thus numerator becomes: $\sin 34^\circ (-\cos 56^\circ) - \sin 56^\circ (\cos 34^\circ) = -(\sin 34^\circ \cos 56^\circ + \sin 56^\circ \cos 34^\circ)$.This matches $-\sin(34^\circ + 56^\circ) = -\sin 90^\circ = -1$.For the denominator: $\cos 178^\circ = \cos(90^\circ + 88^\circ) = -\sin 88^\circ$ and $\sin 208^\circ = \sin(180^\circ + 28^\circ) = -\sin 28^\circ$.Thus denominator becomes: $\cos 28^\circ \cos 88^\circ + (-\sin 88^\circ)(-\sin 28^\circ) = \cos 28^\circ \cos 88^\circ + \sin 88^\circ \sin 28^\circ$.This matches $\cos(88^\circ - 28^\circ) = \cos 60^\circ = \frac{1}{2}$.Now the whole expression is: $\frac{-1}{\frac{1}{2}} = -2$.Answer:A. $-2$