I was reading a textbook and found the following identity for the triangle given in the picture: $$(m+n)\cot(θ)= m\cot(a)-n\cot(b)$$
I tried to prove it using sine and cosine on the triangles but didn't get an expression that seemed to simplify. How should one prove this rule?
[Math] Proof of cotangent rule in a triangle
geometrytrianglestrigonometry
Best Answer
You can prove this rule using the sine law, which gives the following: $$\frac{n}{\sin b} = \frac{q}{\sin (b+\theta)},$$ $$\frac{m}{\sin a} = \frac{q}{\sin (\theta-a)},$$ where $q$ is the line between both triangles. If you now write both equations in terms of $q$, you get the following: $$n \frac{\sin (b+\theta)}{\sin b} = m \frac{\sin (\theta-a)}{\sin a}.$$ Now you can use $\sin (\alpha \pm \beta) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta$ to get: $$n \frac{\sin b \cos \theta + \cos b \sin \theta}{\sin b} = m \frac{\sin \theta \cos a - \cos \theta \sin a}{\sin a}.$$ Simplifying the expression, rearranging the terms and dividing by $\sin \theta$ you get the desired result: $$(m+n) \cot \theta = m \cot a - n \cot b.$$