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

User Jerrod
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1 vote

Answer:

See below.

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.

User Leyton
by
8.1k points

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