# How do I prove this trigonometric identity: (tanΘ+secΘ-1)/(tanΘ-secΘ+... Walk me through it?

The problem didn't show clearly, but it's this: (tanΘ+secΘ-1)/(tanΘ-secΘ+1)=(tanΘ+secΘ)

### 1 Answer

- PuggyLv 71 decade agoFavorite Answer
[tan(t) + sec(t) - 1] / [tan(t) - sec(t) + 1] = tan(t) + sec(t)

To solve this, we choose the more complex side. In this case, it's the left hand side (LHS).

LHS = [tan(t) + sec(t) - 1] / [tan(t) - sec(t) + 1]

Change everything to sines and cosines.

LHS = [sin(t)/cos(t) + 1/cos(t) - 1] / [sin(t)/cos(t) - 1/cos(t) + 1]

Now, multiply everything by cos(t), to get rid of all fractions within fractions.

LHS = [sin(t) + 1 - cos(t)] / [sin(t) - 1 + cos(t)]

Let's rearrange the terms a bit.

LHS = [{sin(t) - cos(t)} + 1] / [{sin(t) + cos(t)} - 1]

Let's multiply the top and bottom by the conjugate of the bottom. That is, [(sin(t) + cos(t)) + 1]. Note that this will lead to a difference of squares in the denominator.

LHS = [{sin(t) - cos(t)} + 1] [(sin(t) + cos(t)) + 1] /

[(sin(t) + cos(t))^2 - 1]

Simplify the denominator by expanding out the squared binomial, to get

LHS = [{sin(t) - cos(t)} + 1] [(sin(t) + cos(t)) + 1] /

[sin^2(t) + 2sin(t)cos(t) + cos^2(t) - 1)

Note that sin^2(t) + cos^2(t) = 1, so

LHS = [{sin(t) - cos(t)} + 1] [(sin(t) + cos(t)) + 1] /

[2sin(t)cos(t) + 1 - 1]

Simplfying some more,

LHS = [{sin(t) - cos(t)} + 1] [(sin(t) + cos(t)) + 1] / [2sin(t)cos(t)]

Now, let's expand the numerator.

LHS = { [sin(t) - cos(t)][sin(t) + cos(t)] + [sin(t) - cos(t)] + [sin(t) + cos(t)] + 1 } / [2sin(t)cos(t)]

Now, we simplify the numerator.

LHS = { sin^2(t) - cos^2(t) + 2sin(t) + 1 } / [2sin(t)cos(t)]

Notice we have a 1, and a -cos^2(t) in the numerator. This translates to 1 - cos^2(t), and can be changed to sin^2(t).

LHS = { sin^2(t) + sin^2(t) + 2sin(t) } / [2sin(t)cos(t)]

Simplify the top,

LHS = [2sin^2(t) + 2sin(t)] / [2sin(t)cos(t)]

Factor the top

LHS = 2sin(t) [sin(t) + 1] / [2sin(t)cos(t)]

The 2sin(t) on the top and bottom cancel.

LHS = (sin(t) + 1) / cos(t)

Now, split this up into two fractions.

LHS = sin(t)/cos(t) + 1/cos(t)

By definition,

LHS = tan(t) + sec(t) = RHS

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