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 = 4θ
- tanθ/4 = θ
Thanks.
Here's one approach:
Use $\displaystyle \tan 2\theta=\frac{2 \tan \theta}{1-\tan^2 \theta}$
$\displaystyle \cos \theta - \tan \theta = 2\cot 2\theta$
$\displaystyle \frac{1}{\tan\theta}-\tan \theta=2 \cot 2\theta$
$\displaystyle \frac{1-\tan^2 \theta}{\tan \theta}=\frac{2}{\tan 2\theta}$
$\displaystyle \frac{1-\tan^2 \theta}{\tan \theta}=\frac{2}{\frac{2 \tan \theta}{1-\tan^2 \theta}}$
$\displaystyle \frac{1-\tan^2 \theta}{\tan \theta}=\frac{1-\tan^2 \theta}{\tan \theta}$
Hello, laoch!
Prove the identity: .$\displaystyle \cot\theta - \tan\theta \:= \:2\cot2\theta$
The right side is: .$\displaystyle \frac{2}{\tan2\theta} \;=\;\frac{2}{\frac{2\tan\theta}{1-\tan^2\!\theta}} \;=\;\frac{2(1-\tan^2\!\theta)}{2\tan\theta}$
. . $\displaystyle =\:\frac{1-\tan^2\!\theta}{\tan\theta} \;= \;\frac{1}{\tan\theta} - \frac{\tan^2\!\theta}{\tan\theta} \;=\; \cot\theta - \tan\theta$