Answer:
See below.
Step-by-step explanation:
I can do this but it's a pretty long proof. There might be a much easier way of proving this but this is the only way I can think of.
Write tan A as s/c and sec A as 1/c ( where s and c are sin A and cos A respectively).
Then tanA+ secA -1/ tanA - secA+1
= (s/c + 1/c - 1) / ( sc - 1/c + 1)
= (s + 1 - c) / c / (s - 1 + c) / c
= (s - c + 1) / (s + c - 1).
Now we write the right side of the identity ( 1 + sin A) / cos A as (1 + s) / c
So if the identity is true then:
(s - c + 1) / (s + c - 1) = (1 + s) / c.
Cross multiplying:
cs - c^2 + c = s + c - 1 + s^2 + cs - s
Simplifying:
cs - c^2 + c = cs - (1 - s^2) + c + s - s
Now the s will disappear on the right side and 1 - s^2 = c^2 so we have
cs - c^2 + c = cs - c^2 + c.
Which completes the proof.
