1. ## Inverse Trigonometric Functions

(1) Show that tan^(-1) v + cot^(-1) v = pi/2

(2) Show that cot^(-1) e^(v) = tan^(-1) e^(-v)

2. Originally Posted by magentarita
(1) Show that tan^(-1) v + cot^(-1) v = pi/2
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)
(2) Show that cot^(-1) e^(v) = tan^(-1) e^(-v)
tan(\theta)= 1/cot(\theta). e^v= 1/e^(-v).

3. ## question 2..........

Originally Posted by HallsofIvy
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).
I can solve question 1 given your tips. Can you solve question 2 step by step?

Thanks