Let $$\alpha$$ and $$\beta$$ be the roots of the quadratic equation $$x^2 \sin\theta - x(\sin\theta \cos\theta + 1) + \cos\theta = 0$$ $$(0 < \theta < 45°)$$, and $$\alpha < \beta$$. Then $$\sum_{n=0}^{\infty}\left(\alpha^n + \frac{(-1)^n}{\beta^n}\right)$$ is equal to:

We start with the quadratic equation

$$x^{2}\sin\theta \;-\;x\bigl(\sin\theta\cos\theta+1\bigr)\;+\;\cos\theta \;=\;0,$$

whose roots are denoted by $$\alpha$$ and $$\beta$$ with $$\alpha<\beta.$$ For a quadratic $$ax^{2}+bx+c=0$$ having roots $$\alpha,\beta,$$ the standard relations

$$\alpha+\beta = -\dfrac{b}{a},\qquad \alpha\beta = \dfrac{c}{a}$$

hold. Comparing term by term we identify

$$a=\sin\theta,$$ $$b=-\bigl(\sin\theta\cos\theta+1\bigr),$$ $$c=\cos\theta.$$

Hence

$$\alpha+\beta =-\dfrac{b}{a} =-\dfrac{-\bigl(\sin\theta\cos\theta+1\bigr)}{\sin\theta} =\dfrac{\sin\theta\cos\theta+1}{\sin\theta} =\cos\theta+\dfrac{1}{\sin\theta},$$

and

$$\alpha\beta =\dfrac{c}{a} =\dfrac{\cos\theta}{\sin\theta} =\cot\theta.$$ Because $$0<\theta<45^\circ,$$ we have $$\cot\theta>1,$$ so $$\alpha\beta>1.$$ That makes one root less than 1 (this will be $$\alpha$$) and the other greater than 1 (this will be $$\beta$$). This fact guarantees the convergence of the two infinite series that will appear next.

We are asked to find

$$S=\sum_{n=0}^{\infty} \left(\alpha^{n}+\dfrac{(-1)^{n}}{\beta^{n}}\right) =\sum_{n=0}^{\infty}\alpha^{n} +\sum_{n=0}^{\infty}\left(\dfrac{-1}{\beta}\right)^{n}.$$ Because $$|\alpha|<1$$ and $$\left|\dfrac{-1}{\beta}\right|<1,$$ each sum is a geometric progression. For a geometric series $$1+r+r^{2}+\dots$$ with common ratio $$|r|<1,$$ the sum is $$\dfrac{1}{1-r}.$$ Applying this formula separately we obtain

$$\sum_{n=0}^{\infty}\alpha^{n} =\dfrac{1}{1-\alpha},$$

and

$$\sum_{n=0}^{\infty}\left(\dfrac{-1}{\beta}\right)^{n} =\dfrac{1}{1-\left(\dfrac{-1}{\beta}\right)} =\dfrac{1}{1+\dfrac{1}{\beta}} =\dfrac{\beta}{\beta+1}.$$

Therefore

$$S =\dfrac{1}{1-\alpha}+\dfrac{\beta}{\beta+1} =\dfrac{1}{1-\alpha}+1-\dfrac{1}{\beta+1}.$$ Re-expressing in a single fraction gives

$$S =1+\left[\dfrac{1}{1-\alpha}-\dfrac{1}{\beta+1}\right] =1+\dfrac{(\beta+1)-(1-\alpha)}{(1-\alpha)(\beta+1)} =1+\dfrac{\alpha+\beta}{(1-\alpha)(\beta+1)}.$$

We now translate everything in the denominator into the symmetric sums already found.

First, compute

$$\bigl(1-\alpha\bigr)\bigl(\beta+1\bigr) =\beta+1-\alpha\beta-\alpha =(\beta-\alpha)+(1-\alpha\beta).$$

To evaluate $$\beta-\alpha$$ we use the difference-of-roots formula

$$\beta-\alpha=\sqrt{(\alpha+\beta)^{2}-4\alpha\beta}.$$

Substituting $$\alpha+\beta=\cos\theta+\dfrac{1}{\sin\theta}$$ and $$\alpha\beta=\dfrac{\cos\theta}{\sin\theta},$$ we have

$$\beta-\alpha =\sqrt{\left(\cos\theta+\dfrac{1}{\sin\theta}\right)^{2} -4\dfrac{\cos\theta}{\sin\theta}} =\sqrt{\cos^{2}\theta+\dfrac{2\cos\theta}{\sin\theta} +\dfrac{1}{\sin^{2}\theta}-\dfrac{4\cos\theta}{\sin\theta}}$$

$$=\sqrt{\cos^{2}\theta -\dfrac{2\cos\theta}{\sin\theta} +\dfrac{1}{\sin^{2}\theta}} =\sqrt{\left(\cos\theta-\dfrac{1}{\sin\theta}\right)^{2}} =\left|\dfrac{1}{\sin\theta}-\cos\theta\right| =\dfrac{1}{\sin\theta}-\cos\theta,$$ since the quantity is positive for $$0<\theta<45^\circ.$$

Also

$$1-\alpha\beta =1-\dfrac{\cos\theta}{\sin\theta} =\dfrac{\sin\theta-\cos\theta}{\sin\theta}.$$

Hence

$$(1-\alpha)(\beta+1) =(\beta-\alpha)+(1-\alpha\beta) =\left(\dfrac{1}{\sin\theta}-\cos\theta\right) +\left(\dfrac{\sin\theta-\cos\theta}{\sin\theta}\right)$$

$$=\dfrac{1-\cos\theta\sin\theta+\sin\theta-\cos\theta}{\sin\theta} =\dfrac{1-\cos\theta\sin\theta+\sin\theta-\cos\theta}{\sin\theta}.$$

For brevity denote this denominator by $$D,$$ so that

$$D =(1-\alpha)(\beta+1) =1-\cos\theta\sin\theta+\sin\theta-\cos\theta.$$

Now the numerator already found is

$$\alpha+\beta =\cos\theta+\dfrac{1}{\sin\theta} =\dfrac{\cos\theta\sin\theta+1}{\sin\theta}.$$

Therefore

$$\dfrac{\alpha+\beta}{(1-\alpha)(\beta+1)} =\dfrac{\dfrac{\cos\theta\sin\theta+1}{\sin\theta}} {1-\cos\theta\sin\theta+\sin\theta-\cos\theta} =\dfrac{\cos\theta\sin\theta+1} {\;1-\cos\theta\sin\theta+\sin\theta-\cos\theta\;}.$$

Adding the 1 that was left outside, we finally get

$$S =1+\dfrac{\cos\theta\sin\theta+1} {1-\cos\theta\sin\theta+\sin\theta-\cos\theta}.$$

To recognise this in the given options, write the right-hand side of Option C, namely $$\dfrac{1}{1-\cos\theta}+\dfrac{1}{1+\sin\theta},$$ as a single fraction:

$$\dfrac{1}{1-\cos\theta}+\dfrac{1}{1+\sin\theta} =\dfrac{(1+\sin\theta)+(1-\cos\theta)} {(1-\cos\theta)(1+\sin\theta)} =\dfrac{2+\sin\theta-\cos\theta} {1-\cos\theta\sin\theta+\sin\theta-\cos\theta}.$$

Observe that

$$1+\dfrac{\cos\theta\sin\theta+1} {1-\cos\theta\sin\theta+\sin\theta-\cos\theta} =\dfrac{\bigl(1-\cos\theta\sin\theta+\sin\theta-\cos\theta\bigr) +\bigl(\cos\theta\sin\theta+1\bigr)} {1-\cos\theta\sin\theta+\sin\theta-\cos\theta} =\dfrac{2+\sin\theta-\cos\theta} {1-\cos\theta\sin\theta+\sin\theta-\cos\theta},$$

exactly the same expression obtained for Option C. Hence the two are identical, so the required infinite sum equals Option C.

Hence, the correct answer is Option 3.