Question 42332
METHOD TO BE FOLLOWED BY SOMEONE WHO IS NEW TO THIS TYPE OF PROBLEMS



First, let us find the quadrant in which the angle {{{-315^o}}} lies.


 -315 = 3 x (-90) -45
(-90) refers to clockwise rotation from axis OX to OY'.
2 x (-90) refers to clockwise rotation from axis OX to OX'.
3 x (-90) refers to clockwise rotation from axis OX to OY.
So, any further clockwise rotation less than {{{90^o}}} keeps the radius vector in the 1st quadrant.


(-45) refers to further {{{45^o}}} clockwise rotation from OY.
See th figure below.
{{{drawing(400,400,-10,10,-10,10,
line(0,0,8,8),
green(line(8,0,8,8)),
grid(1),
locate(8,8,P),
locate(8,0,A),
locate(0,0,O)
)}}}


{{{cot(-315^o)}}} = cot(< AOP) = {{{OA/AP}}} = 1 
[since both OA and AP are along positive x-axis and positive y-axis respectively so both are positive]



METHOD TO BE FOLLOWED BY SOMEONE WHO IS NOT NEW TO THIS TYPE OF PROBLEMS


{{{cot(-315^o)}}}
= {{{cot(360^o-45^o)}}}
= {{{cot(45^o)}}}
= 1