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यदि `A+B=(pi)/(2)`, तो सिद्ध कीजिए कि `cosAcosB` का उच्चतम मान ` (1)/(2)` है।

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माना `x=cosA cos B=(1)/(2)(2cosAcosB)=(1)/(2)[cos(A+B)+cos(A-B)]=(1)/(2)[cos90^(@)+cos(A-B)]`
`=(1)/(2)cos(A-B)`
परन्तु `-1 le cos (A-B)le 1 rArr -(1)/(2)le (1)/(2)cos(A-B)le(1)/(2)`
`rArr -(1)/(2) le x le(1)/(2) rArr -(1)/(2)le cos A cos Ble(1)/(2)`

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