(1) Show that tan^(-1) v + cot^(-1) v = pi/2
(2) Show that cot^(-1) e^(v) = tan^(-1) e^(-v)
Draw a right triangle! tangent is "opposite side over near side" and cotangent is "near side over opposite side" so swapping tan and cot swaps which side is opposite the angle and just swaps the two angles in the right triangle. And the two acute angles in a right triangle always add to pi/2 radians (90 degrees)
tan(\theta)= 1/cot(\theta). e^v= 1/e^(-v).(2) Show that cot^(-1) e^(v) = tan^(-1) e^(-v)