TanA + secA - 1 / tanA - secA+1 = 1 + sinA / cosA

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.