What is the slope of the line that passes through (10, 2) and (6, 10) ?

Refer to the table below if needed. second quadrant third quadrant fourth quadrant sin(1800- - cos(180° -) tan(180°-e) =- tane cot(1800-0) 10 it to solo 888 sin(180° +c) = - sine cos(180° +) =- cose tan(180° +c) = tane cot(180° +o) = cote sec(180° + c) = - seco csc(180° +2) = - csce sin(360° -) =- sine cos(360° -) = cose tan(360° - e) =- tane cot(360° -) = -cote sec(360° -) = seco csc(360° -) = csco sec(180° -) = csc(180° -) = csca 1991 given that sine = 3/5 and lies in quadrant ii, find the following value. tane

2y= 3 2 x−2 y = − 1 3 x + 1 y=− 3 1 x+1

If tan2a=cot(a+60), find the value of a, where 2a is an acute angle