You should use the following trigonometric identity, such that:

`csc alpha = 1/(sin alpha)`

Reasoning by analogy, yields:

`csc (7pi/6) = 1/(sin(7pi/6))`

You may write `(7pi)/6 = pi/2 + (2pi)/3 = (3pi + 4pi)/6` , such that:

`sin(7pi/6) = sin (pi/2 + (2pi)/3) = cos((2pi)/3)`

Replacing `cos((2pi)/3)` for `sin(7pi/6)` yields:

`csc (7pi/6) = 1/(cos((2pi)/3))`

You may use the double angle identity, such that:

`cos 2 alpha = 2cos^2 alpha - 1`

`cos((2pi)/3) = 2 cos^2 (pi/3) - 1`

Replacing `1/2` for `cos (pi/3)` yields:

`cos((2pi)/3) = 2*(1/4) - 1 = 1/2 - 1 = -1/2`

Replacing ` -1/2` for `cos((2pi)/3)` yields:

`csc (7pi/6) = 1/(-1/2) => csc (7pi/6) = -2`

**Hence, evaluating `csc (7pi/6)` , using the indicated trigonometric identities, yields `csc (7pi/6) =-2` .**

`cosec((7pi)/6)=cosec((6pi+pi)/6)`

`=cosec((6pi)/6+pi/6)`

`=cosec(pi+pi/6)`

`=-cosec(pi/6)`

`=(-1)/sin(pi/6)`

`=(-1)/(1/2)`

`=-2`

Note :

`cosec(pi+theta)=-cosec(theta)`

`sin(pi+theta)=-sin(theta)`