Problem in a Trigonometric Series We have to prove that $$\frac{1}{\cos x+\cos3x}+\frac{1}{\cos x+\cos5x}+\frac{1}{\cos x+\cos7x}+...$$ till $n$ terms is:
$$\csc x(\tan(n+1)x-\tan 
 x)$$
Now I've tried solving it by writing applying $\cos a+\cos b$ formula but that didn't lead me anywhere. We definitely have to form a telescoping Series. Now if anyone has any idea how to do so I would really appreciate it.

You started out right by using the formula you have mentioned. Here is a hint to go further:
\begin{align*}
\frac{1}{\cos x + \cos (2n+1)x}&=\frac{1}{2 \cos (n+1)x \,\, \cos nx}\\
&=\frac{\color{red}{\sin x}}{2 \color{red}{\sin x} \cos (n+1)x \,\, \cos nx}\\
&=\frac{\sin [\color{magenta}{(n+1)x-nx}]}{2 \color{red}{\sin x} \cos (n+1)x \,\, \cos nx}
\end{align*}
Now for the numerator use
$$\sin(A-B)=\sin A \cos B- \sin B \cos A,$$
and simplify to get
$$\frac{\csc x}{2}[\tan(n+1)x-\tan nx]$$
then do this for each term and hopefully you will see a pattern.
