<s>[INST] Help with solving mathematical induction problem I need help with the following:
Use mathematical induction to prove that for every $n\in N$, 
$$
\sum_{k=1}^n\frac{1}{\cos kx \cos(k+1)x}=\frac{\tan(n+1)x-\tan x}{\sin x}
$$
For $n=1$, the statement is true.
Suppose that the statement is true for $n=m\in N$, and prove that it is true for $n=m+1$. 
$$\sum_{k=1}^{m+1} \frac{1}{\cos kx \cos(k+1)x}=\frac{\tan(m+2)x-\tan x}{\sin x}(*)$$
Proof:
$$\frac{\tan(m+2)x-\tan x}{\sin x}=\frac{\tan(m+1)x-\tan x}{\sin x}+\frac{1}{\cos(k+1)x \cos(k+2)x}$$
Using trigonometry rule $\cos x \cos y$, the right side of equation is
$$\frac{(\cos x+\cos(2k+3)x)(\tan(m+1)x-\tan x)+2\sin x}{\sin x(\cos x+\cos(2k+3)x)}$$
How to further transform this expression to become $(*)$
Thanks for replies.
 [/INST] $$\frac{\tan(m+2)x-\tan x}{\sin x}-\frac{\tan(m+1)x-\tan x}{\sin x}$$
$$=\frac{\tan(m+2)x-\tan(m+1)x}{\sin x}$$
$$=\dfrac{\sin[(m+2)x-\sin(m+1)x]}{\cos(m+2)x\cdot\cos(m+1)x\cdot \sin x}$$
$$=\dfrac1{\cos(m+2)x\cdot\cos(m+1)x}$$ if $\sin x\ne0$
</s>