Question

`(tan) (11pi)/3 (-2 sin) (4pi)/6 - 3/4 (cosec^2) pi/4 +(4cos^2) (17pi)/6 = (3-4sqrt3)/2`

Answer 1

`tan((11pi)/3) = tan((9pi+2pi)/3) = tan(3pi+(2pi)/3) = tan(pi+(2pi)/3) = tan((2pi)/3) = -sqrt3`
`sin((4pi)/6) = sin((2pi)/3) = sqrt3/2`
`cosec^2(pi/4) = (sqrt 2)^2 = 2`
`cos^2((17pi)/6) = cos^2((18pi-pi)/6) = cos^2(3pi-pi/6) = cos^2(pi-pi/6) = (-sqrt3/2)^2 = 3/4`
So, putting these values in the given expression,
`-sqrt3-2*sqrt3/2-3/4*2+4*3/4 = -2sqrt3+3/2 = 1/2(3-4sqrt3) `
which is the desired value.