3cos²π/6 + 3cosec²π/3 . secπ/3+sin²4π/3 kindly solve this . Created by shubhramalvankar2007.

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October 2023 1 6 Report

3cos²π/6 + 3cosec²π/3 . secπ/3+sin²4π/3

kindly solve this

3cos²30 + 3cosec²60 . sec 60 + sin²240

3cos²30 + 3cosec²60 . sec 60 + sin²(180 + 60)

3cos²30 + 3cosec²60 . sec 60 - sin²60

[ 3 × ( √3 / 2 )² ]+ [ 3 × ( 2 / √3 )² . 2 ] - ( √3 / 2 )²

= 3 ( 3 / 4 ) + 3 ( 4 / 3 ) .2 - ( 3 / 4 )

= 9 / 4 + 8 - 3 / 4

= 9 + 32 - 3 / 4

= 38 / 4

= 19 / 2