Question 851835
Since tan and sec is negative therefore it is in quadrant 2, wherein only sin and csc are positive. :)

sec = -8/5 therefore,
cos = -5/8 because cos is the reciprocal of sec,

It is stated in the Pythagorean identities that,
cos^2+sin^2 =1

Therefore,
sin^2 =1-cos^2

Substitute,

sin^2 =1-(-5/8)^2
sin^2 =1-(25/64)
sin^2 =(64/64)-(25/64)
sqrt(sin^2) = sqrt(39/64)
sin = sqrt39/8

That is the value of sin squareroot of 39 over 8, hope that helps! :)