# Trigonometric Identity Problem

1. Nov 4, 2012

### KingKai

1. The problem statement, all variables and given/known data
Prove the Identity

sinθ/(1+cosθ) = 1-cos(θ)/sinθ

2. Relevant equations

sinθ/cosθ = tanθ

sin^2θ + cos^2θ = 1

3. The attempt at a solution

sinθ/(1 + cosθ) = LS

cosθtanθ/(1+cosθ) = LS

cosθtanθ/(sin^2θ + cos^2θ + cosθ) = LS

cosθtanθ/(tan^2θcos^2θ + cos^2θ + cosθ) = LS

(cosθ/cosθ) (tanθ/[tan^2θcosθ + cosθ + 1]) = LS

sinθ/(cosθ[tan^2θcosθ + cosθ + 1]) = LS

sinθ/(sin^2θ + cos^2θ +1) = LS

sinθ/2 = LS

Last edited by a moderator: Nov 4, 2012
2. Nov 4, 2012

### SammyS

Staff Emeritus
The above line should be: $\displaystyle \frac{\cos(\theta)}{\cos(\theta)}\left( \frac{\sin(\theta)}{\tan^2(\theta)\cos(\theta) + \cos(\theta) + 1}\right)$
What is  1 - cos2(θ) ?

How can you make the denominator of LHS or numerator of RHS equal to  1 - cos2(θ) ?

Last edited by a moderator: Nov 4, 2012
3. Nov 4, 2012

### KingKai

why should sinθ be in the numerator?

I factored out (cosθ/cosθ), where the numerator was initially

cosθtanθ

thus leaving me with tanθ instead of sinθ in the numerator.

4. Nov 4, 2012

### SammyS

Staff Emeritus
$\displaystyle \cos(\theta)\tan(\theta)=\cos(\theta)\frac{\sin( \theta)}{\cos(\theta)}$
$\displaystyle =\frac{\cos(\theta)\,\sin(\theta)}{\cos(\theta)}$

$\displaystyle =\sin(\theta)$​

5. Nov 4, 2012

### Villyer

In those steps you didn't distribute the cos back to the 1.

Last edited by a moderator: Nov 4, 2012
6. Nov 4, 2012

### ehild

It is not an identity. sinθ/(1+cosθ) = (1-cos(θ))/sinθ is.

ehild

7. Nov 5, 2012

### SammyS

Staff Emeritus
Good point !

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