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Question Number 216836 by hardmath last updated on 22/Feb/25

$$\mathrm{Find}:$$$$\frac{(1+\tan 1°)(1+\tan 2°)...(1+\tan 44°)}{(1-\tan 46°)(1-\tan 47°)...(1-\tan 89°)}=?$$

Answered by BaliramKumar last updated on 22/Feb/25

$$1$$

Answered by mehdee7396 last updated on 22/Feb/25

$$\frac{(s1+s89)(s2+s88)....(s43+s47)(s44+s46)}{(s44-s46)(s43-s47)...(s2-s88)(s1-s89)}\times \frac{c46\times c47\times ...\times c89}{c1\times c2\times ...\times c44}$$$$=\frac{s45c44}{-s45c1}\times \frac{s45c43}{-s45c2}\times \frac{s45c42}{-s45c3}\times ...\times \frac{s45c2}{-s45c43}\times \frac{s45c1}{-s45c44}$$$$=1$$$$$$

Commented by hardmath last updated on 22/Feb/25

$$\mathrm{s}=\mathrm{sinus}$$$$\mathrm{c}=\mathrm{cosinus}$$

Answered by MrGaster last updated on 23/Feb/25

$$\mathrm{Let}P=\underset{k=1}{\prod ^{44}}(1+\tan k°)\wedge Q=\underset{k=46}{\prod ^{89}}(1-\tan k°).\mathrm{Find}\left(\frac{P}{Q}\right)$$$$\tan (90°-x°)=\cot x°⊢\tan ({90}^{°}-x°)=\cot x°,\mathrm{we}\mathrm{have}:$$$$\tan (90°-k°)=\cot k°=\frac{1}{\tan k°}$$$$\Rightarrow 1-\tan (90°-k)=1-\frac{1}{\tan k°}=\frac{\tan k°-1}{\tan k°}$$$$\mathrm{Now}\mathrm{consider}\mathrm{the}\mathrm{product}:$$$$P\cdot \underset{k=1}{\prod ^{44}}\frac{\tan k°-1}{\tan k°}=\underset{k=1}{\prod ^{44}}(1+\tan k°)\cdot \frac{\tan k°-1}{\tan k°}=\underset{k=1}{\prod ^{44}}\frac{(\tan k°+1)(\tan k°-1)}{\tan k°}$$$$=\underset{k=1}{\prod ^{44}}\frac{{\tan }^{2}k°-1}{\tan k°}$$$$\tan (45°+x°)=\frac{1+\tan x°}{1-\tan x°}⊢\tan ({45}^{°}+k°)\cdot \tan (45°-k°)=1$$$$\Rightarrow \tan (45°+k°)=\frac{1}{\tan (45°-k°)}$$$$\mathrm{This}\mathrm{implies}:$$$$\tan (45°+k°)\cdot \tan (45\cdot k°)=\frac{1+\tan k°}{1-\tan k}\cdot \frac{1-\tan k}{1+\tan k}=1$$$$\mathrm{so},\mathrm{\Pi }\mathrm{simplifies}\mathrm{to}:$$$$P\cdot \underset{k=1}{\prod ^{44}}\frac{\tan k°-1}{\tan k°}=\underset{k=1}{\prod ^{44}}\frac{{\tan }^{2}k°-1}{\tan k°}=\underset{k=1}{\prod ^{44}}\tan ({45}^{°}+k°)\cdot \tan (45°-k°)$$$$=\underset{k=1}{\prod ^{44}}1=1$$$$\mathrm{so},P=\underset{k=1}{\prod ^{44}}\frac{\tan k°}{\tan k°-1},\mathrm{Now}\mathrm{consider}:$$$$\frac{P}{Q}=\frac{\underset{k=1}{\prod ^{44}}(1+\tan k°)}{\underset{k=46}{\prod ^{89}}(1-\tan k°)}=\underset{k=1}{\prod ^{44}}\frac{1+\tan k°}{1-\tan (90°-k°)}$$$$=\underset{k=1}{\prod ^{44}}\frac{1+\tan k°}{1-\cot k°}=\underset{k=1}{\prod ^{44}}\frac{1+\tan k°}{\frac{1-\tan k°}{\tan k°}}$$$$=\underset{k=1}{\prod ^{44}}\frac{\tan k°(1+\tan k°)}{1-\tan k°}$$$$\tan (45°+x°)=\frac{1+\tan x°}{1-\tan x°}⊢\tan (45°+k^{°})=\frac{1+\tan k°}{1-\tan k°}$$$$\Rightarrow \frac{P}{Q}=\underset{k=1}{\prod ^{44}}\tan (45°+k°)=\tan (45°+1°)\cdot \tan (45+°2)\cdot \dots \cdot \tan (45°+44°)$$$$\because \tan (45°+x°)=\frac{1+\tan x°}{1-\tan x}\Rightarrow \frac{P}{Q}=\frac{1+\tan 1}{1-\tan 1°}\cdot \frac{1+\tan 2°}{1-\tan 2°}\cdot \dots \cdot \frac{1+\tan 44°}{1-\tan 44°}$$$$\mathrm{so},=\overline{)\begin{array}{c}1\end{array}}$$

Commented by hardmath last updated on 23/Feb/25

$$$$Excellent solution, thank you very much dear professor

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