Angle between two lines without using the tan identity as an exercies to my self I wanted to proof that the well known formula $$\tan(\theta)=\frac{m_1- m_2}{1+m_1\times m_2}$$where $m_1$ and $m_2$ are the respectivei gradients,
for the angle between two lines. I tried proving it through vector geometry with no success. Then I realised that $\tan(\theta)=m_1$ and $\tan(\theta)=\frac{1}{m_2}$. Then from the RHS:
$$=\frac{\tan\theta-\frac{1}{\tan\theta}}{1+1}$$
$$=\frac{\sin^2\theta-\cos^2\theta}{2\cos\theta\sin\theta}$$
$$=-\cot2\theta$$
could somebody please explain to me where I went wrong?
 A: The sentence 'Then, I realize' is not correct. Suppose  $$\tan(\theta)= m_1$$
$$\tan(\beta) =m_2$$. Note that $ \theta -\beta$ is the angle between the given lines. I hope this helps. 
In fact, the following trig identity and the gradient identity you've written are the same things. 
Recall that the slope/ gradient/rate of change/ derivative of a straight line is $$\tan\theta$$ where $\theta$ is the angle between the given straight line and $\mathbb{X}$-axis.
$$ \tan(\theta - \beta)=\frac{(\tan\theta - \tan\beta)}{1+ \tan\theta\times \tan\beta}$$
A: Note $m_1=\tan\theta_1=\frac{\sin\theta_1}{\cos\theta_1}$ and $m_2=\tan\theta_2=\frac{\sin\theta_2}{\cos\theta_2}$. Then, $\theta=\theta_1-\theta_2$ and
$$\frac{m_1- m_2}{1+m_1\cdot m_2}=\frac{\frac{\sin\theta_1}{\cos\theta_1}-\frac{\sin\theta_2}{\cos\theta_2}}{1+\frac{\sin\theta_1}{\cos\theta_1}\frac{\sin\theta_2}{\cos\theta_2}}=\frac{\sin\theta_1\cos\theta_2-\cos\theta_1\sin\theta_2}{\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2}=\frac{\sin(\theta_1-\theta_2)}{\cos(\theta_1-\theta_2)}=\tan\theta$$
