The least value of 8 cosec2θ + 25 sin2 θ is: A10210\sqrt{2}102B40240\sqrt{2}402C20220\sqrt{2}202D30230\sqrt{2}302Answer: 20220\sqrt{2}202 Read Explanation: Solution:Given:8 cosec2θ + 25 sin2 θFormula:The minimum value of "a cosec2θ + b sin2 θ" is given by 2ab2\sqrt{ab}2abCalculation:Minimum Value of 8 cosec2θ + 25 sin2 θ =2(8×25)=2\sqrt{(8\times{25}})=2(8×25)=2200=2\sqrt{200}=2200=2(100×2)=2\sqrt{(100\times{2}})=2(100×2)=2×102=2\times{10}\sqrt{2}=2×102=202=20\sqrt{2}=202 Read more in App