Question #758df

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Answer 1 · 2017-04-17T15:19:28

$tan((11pi)/3)-2sin((4pi)/6)-(3/4)cosec^2(pi/4)+4cos^2((17pi)/6)$

$=tan(4pi-pi/3)-2sin(pi-pi/3)-(3/4)cosec^2(pi/4)+4cos^2(3pi-pi/6)$

$=-tan(pi/3)-2sin(pi/3)-(3/4)cosec^2(pi/4)+4cos^2(pi/6)$

$=-sqrt3-2xxsqrt3/2-(3/4)xx(sqrt2)^2+4(sqrt3/2)^2$

$=-sqrt3-sqrt3-(3/4)xx2+4xx3/4$

$=-sqrt3-sqrt3-3/2+3$

$=3/2-2sqrt3$