Question

4/3 cot²30° + 3 sin²30° - 2 cosec²60° - 3/4 tan²30°

karo Yaar koi to solve....​

Answer 1

43cot230+3sin260−2cosec260−34tan230Puttingthevaluesofsin60,cot30,cosec60tan30weget=43(3)−−√2+3(3√2)2−2(23√)2−34(13√)2=4+94−83−14=4+2−83=6−83=(18−8)3=103

Answer 2

Answer:

answer for the given problem is given