(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