Answer: (a) \(183\) approximately. (b) The sum to infinity is \(\operatorname{cosec}\theta+\cot\theta\).

(a) Let the arithmetic progression have \(n\) terms and common difference \(d\).

Using the sum formula,

\(\frac n2(9+159)=2604.\)

So

\(84n=2604,\)

and hence

\(n=31.\)

The last term is

\(159=9+30d,\)

so

\(d=5.\)

The \(12\)th term of the arithmetic progression is

\(9+11(5)=64.\)

This is the first term of the geometric progression.

The \(8\)th term of the arithmetic progression is

\(9+7(5)=44.\)

This is the second term of the geometric progression. Therefore the common ratio is

\(r=\frac{44}{64}=\frac{11}{16}.\)

The sum of the first \(6\) terms is

\(S_6=\frac{64\left(1-\left(\frac{11}{16}\right)^6\right)}{1-\frac{11}{16}}=183.17\ldots.\)

So the sum is approximately \(183\).

(b)(i) The common ratio is \(\cos\theta\), where \(45^\circ\leqslant\theta\leqslant135^\circ\).

Therefore

\(-\frac{\sqrt2}{2}\leqslant\cos\theta\leqslant\frac{\sqrt2}{2}.\)

Hence \(|\cos\theta|\lt 1\), so the geometric progression has a sum to infinity.

(b)(ii) The sum to infinity is

\(\frac{a}{1-r}=\frac{\sin\theta}{1-\cos\theta}.\)

Rationalise the denominator:

\(\frac{\sin\theta}{1-\cos\theta}\cdot\frac{1+\cos\theta}{1+\cos\theta}=\frac{\sin\theta(1+\cos\theta)}{1-\cos^2\theta}.\)

Since \(1-\cos^2\theta=\sin^2\theta\),

\(\frac{\sin\theta(1+\cos\theta)}{\sin^2\theta}=\frac{1+\cos\theta}{\sin\theta}.\)

Thus

\(\frac{1+\cos\theta}{\sin\theta}=\operatorname{cosec}\theta+\cot\theta.\)