1. Prove the identity

Hi again,

I would appreciate it a lot if anyone could tell me if this is correct:

Prove the identity:
cotθ - tanθ = 2cot2θ

cosθ/sinθ - sinθ/cosθ = 2(cosθ/sinθ)2θ
- sinθ/cosθ =
- tanθ/4 = θ

Thanks.

2. Originally Posted by laoch
Hi again,

I would appreciate it a lot if anyone could tell me if this is correct:

Prove the identity:
cotθ - tanθ = 2cot2θ

cosθ/sinθ - sinθ/cosθ = 2(cosθ/sinθ)2θ
- sinθ/cosθ =
- tanθ/4 = θ

Thanks.
Sorry, what?

Anyway here is my proof.
cot T - tan T = 2 cot 2T
iff cos T / sin T - sin T / cos T = 2 cos 2T / sin 2 T
iff (cos^2 T - sin^2 T) / sin T cos T = 2 (cos^2 T - sin^T) / 2 sin T cos T
QED.
Identities used: cos 2T = cos^2 T - sin^2 T and sin 2T = 2 sin T cos T

3. Originally Posted by laoch
Hi again,

I would appreciate it a lot if anyone could tell me if this is correct:

Prove the identity:
cotθ - tanθ = 2cot2θ

cosθ/sinθ - sinθ/cosθ = 2(cosθ/sinθ)2θ
- sinθ/cosθ =
- tanθ/4 = θ

Thanks.
Here's one approach:

Use $\tan 2\theta=\frac{2 \tan \theta}{1-\tan^2 \theta}$

$\cos \theta - \tan \theta = 2\cot 2\theta$

$\frac{1}{\tan\theta}-\tan \theta=2 \cot 2\theta$

$\frac{1-\tan^2 \theta}{\tan \theta}=\frac{2}{\tan 2\theta}$

$\frac{1-\tan^2 \theta}{\tan \theta}=\frac{2}{\frac{2 \tan \theta}{1-\tan^2 \theta}}$

$\frac{1-\tan^2 \theta}{\tan \theta}=\frac{1-\tan^2 \theta}{\tan \theta}$

4. Hello, laoch!

Prove the identity: . $\cot\theta - \tan\theta \:= \:2\cot2\theta$

The right side is: . $\frac{2}{\tan2\theta} \;=\;\frac{2}{\frac{2\tan\theta}{1-\tan^2\!\theta}} \;=\;\frac{2(1-\tan^2\!\theta)}{2\tan\theta}$

. . $=\:\frac{1-\tan^2\!\theta}{\tan\theta} \;= \;\frac{1}{\tan\theta} - \frac{\tan^2\!\theta}{\tan\theta} \;=\; \cot\theta - \tan\theta$