UPSC NDA I 2019 Mathematics Solved Paper

(D) Given expression Δ = 25 cosec‌2 x + 36 sec‌2 x
⇒ Δ = 25(1+cot‌2 x ) + 36(1+tan‌2 x )
Δ = 61 + 25cot‌2 x + 36 tan‌2 x
For minimum value of 25 cot‌2× + 36 tan‌2x
applying AM−GM inequality

25cot2x+36tan2x

2

≥ (25cot2x.36tan2x)1∕2
⇒ 25cot‌2x+36 tan‌2x ≥ 60
⇒ 61+25cot‌2x+36 tan‌2x ≥121
Hence minimum value of
Δ = 121