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### Proof of a Trig Identity

```
Date: 03/13/98 at 20:03:02
From: Chris Hebert
Subject: Proving Trigonomic Identies

How can I prove that

1 + sinX + cosX      1 + cosX
------------------- = -----------
1 + sinX - cosX        sinX

I have tried solving this problem using

Basic Trigonometric Identities,
Pythagorean Identities,
Double-Angle Formulas, and
Half-Angle Formulas.

This problem has stumped the entire Pre-Calculus class, including my
teacher! Any help would be appreciated. Even if this problem stumps
you, will you please E-mail and tell me so, or let me know any
progress you have made? (I do know that this problem is solvable
though, because it was published on a work sheet).
```

```
Date: 03/16/98 at 17:14:13
From: Doctor Fred
Subject: Re: Proving Trigonomic Identies

Proof:
Often, multiplying by conjugates is helpful in trig identities.

1 + sinX + cosX           (1 + sinX + cosX)^2
---------------- =  -----------------------------------
1 + sinX - cosX      (1 + sinX - cosX)(1 + sinX + cosX)

1 + 2sinX + 2cosX + 2sinXcosX + (sinX)^2 + (cosX)^2
= -----------------------------------------------------
1 + 2sinX + (sinX)^2 - (cosX)^2

Now, notice almost everything has a coefficient of 2. Substituting 1
for (sinX)^2 + (cosX)^2 in the numerator and (sinX)^2 for 1 - (cosX)^2
in the denominator gives

2 + 2sinX + 2cosX + 2sinXcosX
= -------------------------------
2sinX + 2(sinX)^2

Factor by grouping in the numerator, and factor out sinX in the
denominator

(1 + sinX)(1 + cosX)
= ----------------------
(1 + sinX)sinX

Dividing numerator and denominator by 1+sinX gives

1 + cosX
= -----------
sinX

-Doctor Fred, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Trigonometry

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