Give line is
5x−12y+6=0Slope of this line is
m1=‌=‌Since, line
L is perpendicular to this line, the product of their slope is -1 . Let the slope of line
L be
mL.
Then,
mL×‌=−1⇒‌‌mL=−‌The equation of a line in normal form is
x‌cos‌θ+ysin‌θ=P, where
P is the distance from the origin to the line and
θ is the angle made by the perpendicular from the origin.
We have
P=2 unit, So, the equation of line is
x‌cos‌θ+ysin‌θ=2Slope,
mL=‌=−cot‌θ∴−cot‌θ=‌⇒cot‌θ=‌For
y-intercept,
x=0 then
ysin‌θ=2⇒‌‌y=‌And since
y-intercept is positive, so
‌‌>0‌⇒sin‌θ>0⇒cot‌θ=‌>0,θ must be in first quadrant.
Now,
tan‌θ=‌=‌So,
tan‌θ+cot‌θ=‌+‌=‌=‌