UPSC NDA I 2019 Mathematics Solved Paper

(A)
sin‌2(α−β ) + 2abcos(α−β)
= 1 − cos‌2( α−β ) + 2 ab cos(α−β )
=1 − cos(α−β ) [ cos( α−β)−2ab]
= 1 − cos(α−β ) [√1−a2√1−b2 −ab]
= 1−(√1−a2√1−b2 +ab)(√1−a2√1−b2 −ab)
= 1−[(1−a2) (1−b2) − a2b2 ]
= 1 − [ 1 − a2−b2 + a2b2 − a2b2]
= a2+b2
∴ sin‌2 (α−β ) + 2ab cos(α−β ) = a2+b2 √1−a2