Differentiate logarithmically the expressions for $\sin{\theta}$ and $\cos{\theta}$ in factors and deduce the sums to infinity. (i)$$\frac{1}{ \theta^2 - \pi^2} + \frac{1}{ \theta^2- 2^2\pi^2} + \frac{1}{ \theta-  3^2\pi^2 }... $$
(ii)$$\frac{1}{1^2 + x^2} + \frac{1}{2^2 +x^2} + \frac{1}{3^2 + x^2}... $$
These are the problems given in "Differential Calculus for Beginners"" by Joseph Edwards. I am beginner and have no idea how to approach such problems. The phrase 'differentiate logarithmically' is confusing me.

The phrase "the expressions for $\sin\theta$ and $\cos\theta$ in factors" seems to refer to the infinite product representations of those functions.
For (i):
\begin{align*}
\sin\theta &=
\theta \; \prod_{n=1}^{\infty}\left(1 - \frac{\theta^2}{n^2 \pi^2}\right)\\
\frac{d}{d\theta}\ln(\sin\theta) &=
\frac{d}{d\theta}\left[\ln\theta \;+\; \sum_{n=1}^{\infty}\ln\left(1-\frac{\theta^2}{n^2\pi^2}\right)\right]\\
&=
\frac{1}{\theta} \;+\; 2\theta \sum_{n=1}^{\infty}\frac{1}{\theta^2 - n^2\pi^2}
\end{align*}
From there you can solve for the sum in question, to yield:
$$
\sum_{n=1}^{\infty}\frac{1}{\theta^2 - n^2\pi^2} \;=\; 
\frac{1}{2\theta}\left(\cot\theta \;-\; \frac{1}{\theta}\right)
$$
For (ii), the simplest thing to do is to use the above result with $\theta\rightarrow i \pi x$.
