Sin/Cos=Tan

From

$\begin{aligned} \tan^{a_i} (45^\circ - i^\circ) & = \tan (44^\circ - i^\circ) \\ a_i \log \left(\tan (45^\circ - i^\circ) \right) & = \log \left(\tan (44^\circ - i^\circ) \right) \\ \implies a_i & = \frac {\log \left(\tan (44^\circ - i^\circ) \right)}{\log \left(\tan (45^\circ - i^\circ) \right)} \\ & = \frac {\log \left(\tan (45^\circ - (i+1)^\circ) \right)}{\log \left(\tan (45^\circ - i^\circ) \right)} \\ & = \frac {\log \left(\frac {1-\tan(i+1)^\circ}{1+\tan(i+1)^\circ} \right)}{\log \left(\frac {1-\tan i^\circ}{1+\tan i^\circ} \right)} \\ & = \frac {\log \left(1-\tan(i+1)^\circ \right) - \log \left(1+\tan(i+1)^\circ \right)}{\log \left(1-\tan i^\circ \right) - \log \left(1+\tan i^\circ \right)} \end{aligned}$

Similarly,

$\begin{aligned} \implies b_i & = \frac {\log \left(\tan (46^\circ + i^\circ) \right)}{\log \left(\tan (45^\circ + i^\circ) \right)} \\ & = \frac {\log \left(\tan (45^\circ + (i+1)^\circ) \right)}{\log \left(\tan (45^\circ + i^\circ) \right)} \\ & = \frac {\log \left(\frac {1+\tan(i+1)^\circ}{1-\tan(i+1)^\circ} \right)}{\log \left(\frac {1+\tan i^\circ}{1-\tan i^\circ} \right)} \\ & = \frac {\log \left(1+\tan(i+1)^\circ \right) - \log \left(1-\tan(i+1)^\circ \right)}{\log \left(1+\tan i^\circ \right) - \log \left(1-\tan i^\circ \right)} \end{aligned}$

We note that

$\begin{aligned} \frac {\log \left(1-\tan(i+1)^\circ \right) - \log \left(1+\tan(i+1)^\circ \right)}{\log \left(1-\tan i^\circ \right) - \log \left(1+\tan i^\circ \right)} & = \frac {\log \left(1+\tan(i+1)^\circ \right) - \log \left(1-\tan(i+1)^\circ \right)}{\log \left(1+\tan i^\circ \right) - \log \left(1-\tan i^\circ \right)} \\ \implies a_i & = b_i \\ \frac {a_i}{b_i} & = 1 \quad \quad \small \color{#3D99F6} \text{Note that } a_i = b_i > 0 \\ \implies \frac {a_1a_2a_3...a_{43}}{b_1b_2b_3...b_{43}} & = \boxed{1} \end{aligned}$

$\large\begin{aligned}\begin{cases}\tan^{a_{i}}(45^\circ-i^\circ)&=\tan(44^\circ-i^\circ)\hspace {5mm}\\ \large\tan^{b_{i}}(45^\circ+i^\circ)&=\tan(46^\circ+i^\circ)\color{#3D99F6}\end{cases}\color{#3D99F6}\small\text{Given}\\\\\end{aligned}$ $\color{#333333}\large\begin{aligned}\text{Now we have,}\\\\ \tan^{b_{i}}(45^\circ+i^\circ)&=\tan(46^\circ+i^\circ)\\\\ &=\cot(44^\circ-i^\circ)\hspace{5mm}\small\color{#3D99F6}{(46+i)+(44-i)=90},\tan(90^\circ-i^\circ)=\cot(i^\circ)\\\\ &=\dfrac{1}{\tan(44^\circ-i^\circ)}\\\\ &=\dfrac{1}{\tan^{a_{i}}(45^\circ-i^\circ)}\hspace{5mm}\small\color{#3D99F6}\tan(44^\circ-i^\circ)=\tan^{a_{i}}(45^\circ-i^\circ)\\\\ \tan^{b_{i}}(45^\circ+i^\circ)&=\cot^{a_{i}}(45^\circ-i^\circ)\\\\ \tan^{b_{i}}(45^\circ+i^\circ)&=\tan^{a_{i}}(45^\circ+i^\circ)\hspace{5mm}\small\color{#3D99F6}(45+i)+(45-i)=90,\cot(90^\circ-i^\circ)=\tan(i^\circ)\\\\ \implies {b_{i}}&={a_{i}}\hspace{5mm}\small\color{#3D99F6}\text{As,} 1<\tan(45^\circ+i^\circ)<\infty\\\\ \implies\dfrac{a_{i}}{b_{i}}&=1\\ \prod_{i=1}^{43}\dfrac{a_{i}}{b_{i}}&=1^{43}=\color{#EC7300}\boxed{{\color{#333333}1}}\end{aligned}$