#sec(arctan(-3/5))#
Let's work with the first part: #arctan(-3/5)#
The domain for #arctan# is limited to the first and fourth quadrants. Because the lengths are negative, the triangle must lie in the #4th# quadrant.
tangent corresponds to #(opp)/(adj)#:
#color(white)(--)5#
#color(white)(.)color(black)(----)#
# color(white)(x)color(black)(\\\\)color(white)(--......)|#
#color(white)(x)color(white)(.)color(black)(\\\\)color(white)(---)|#
#color(white)(x)color(white)(..)color(black)(\\\\)color(white)(-.......)|#
#color(black)(x)color(white)(....)color(black)(\\\\)color(white)(-.... .)|#
#color(white)(x)color(white)(.....)color(black)(\\\\)color(white)(-....)|##-3#
#color(white)(x)color(white)(.......)color(black)(\\\\)color(white)(-..)|#
#color(white)(x)color(white)(........)color(black)(\\\\)color(white)(-)|#
#color(white)(x)color(white)(..........)color(black)(\\\\)color(white)(..)|#
#color(white)(x)color(white)(...........)color(black)(\\\\)color(white)(.)|#
#color(white)(x)color(white)(.............)color(black)(\\\\)color(white)()|#
We have this picture. Using it, we need to find #sec#. To do that, we need #(hyp)/(adj)#.
#adj# is #5#, but to find the #hyp#, we need to use pythagorean's theorem: #sqrt((-3)^+5^2)=sqrt(34)#.
Now we have all our info: #sec=(hyp)/(adj)=(sqrt(34))/5#
