If csc (weird squiggly thing) = 13/5 then what would sec sin cos and tan = ?,

Given cosec x = 13/5, x being the angle. cosec x = 1 / Sin x = > sin x = 1 / cosec x = > sin x = 1 / (13/5) = > sin x = 5/13 So in a right angled triangle sin x = perpendicular height (h) / l, l being the side opposite to right angle. So h = 5, l = 13. From Pythagoras theorem l^2 = h^2 + b^2, b is the base 13^2 = 5^2 + b^2 = > b^2 = 13^2 - 5^2 = > b^2 = 169 - 25 = > b^2 = 144 = > b = 12 sin x = 5/13 cos x = b / l = 12/13 sec x = 1/cos x = 1 / (12/13) = 13/12 tan x = sin x / cos x = (5/13) / (12/13) = 5/12