Question

How do you find the value of `sin ((2pi)/3) Cos pi Tan ((2pi)/3)Cot ((5pi)/6)`?

Answer 1

`=>sin ((2pi)/3) Cos pi Tan ((2pi)/3)Cot ((5pi)/6)`
`=1/2 * -1 * -1/sqrt3*-1/sqrt3 =-1/6`

Explanation:

`cos pi = cos180 = -1`

`sin((2pi)/3)= sin 120 = sin (90 +30) = sin 30 = 1/2`

`Tan ((2pi)/3) = tan 120 = -tan 30 = -1/sqrt3`

`Cot ((5pi)/6) = cot 150 = - cot 60 = -tan(90-60)`
`= -tan30 =- 1/sqrt3`

`=>sin ((2pi)/3) Cos pi Tan ((2pi)/3)Cot ((5pi)/6)`
`=1/2 * -1 * -1/sqrt3*-1/sqrt3 =-1/6`