Question: The question is let $\{m\_1, m\_2, m\_3, \dots \}$ be a sequence of numbers where $m\_k\geq 0$ for every $k \geq 1$.

Let $$M\_n = \sum\_{k=1}^n m\_k $$ when $n \geq 1$ is an integer.

Show that if $$\prod\_{k=1}^{n-1}{(M\_k+1)\choose(M\_k)}= \frac{(m\_1+m\_2+\cdots+m\_n)!}{(m\_1!m\_2!\cdots m\_n!)}$$

When I try I get an expansion that is for example the first term and I choose a random letter in this case 2 for $m\_1$ the number $2!+1\over 2!(2+1-2)!$.

Is this the correct way to expand this series?

Thanks in advance. Let me know if anything should be clarified.

Answer: 
**Hint**

Assuming that the equation is $$\tan(3x) - \cot(4x)=0$$ rewrite it as $$\frac{\sin(3x)}{\cos(3x)}-\frac{\cos(4x)}{\sin(4x)}=\frac{\sin(3x)\sin(4x)-\cos(3x)\cos(4x)}{\cos(3x)\sin(4x)}=-\frac{\cos(7x)}{\cos(3x)\sin(4x)}=0$$

I am sure that you can take from here.
If you can use the complex exponential definition of trig functions,

$$\sin(\theta) = \frac{e^{i\theta} - e^{-i\theta}}{2i}$$
$$\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$$
$$\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{(e^{i\theta} - e^{-i\theta})}{i(e^{i\theta} + e^{-i\theta})}$$
$$\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)} = \frac{i(e^{i\theta} + e^{-i\theta})}{(e^{i\theta} - e^{-i\theta})}$$

Then your equation becomes:

$$\frac{(e^{i3x} - e^{-i3x})}{i(e^{i3x} + e^{-i3x})} = \frac{i(e^{i4x} + e^{-i4x})}{(e^{i4x} - e^{-i4x})}$$
$$(e^{i3x} - e^{-i3x})(e^{i4x} - e^{-i4x}) = -(e^{i3x} + e^{-i3x})(e^{i4x} + e^{-i4x})$$
$$e^{i7x} - e^{-ix} - e^{ix} + e^{-i7x} = -(e^{i7x} + e^{-ix} + e^{ix} + e^{-i7x})$$
$$2e^{i7x} + 2e^{-i7x} = 0$$
$$2(2\cos(7x)) = 0$$
$$\cos(7x) = 0$$
$$7x = \frac{\pi}{2} + \pi n, n \in \mathbb{Z}$$
$$x = \frac{\pi}{14} + \frac{\pi}{7}n$$