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Question Number 191144 by mathlove last updated on 19/Apr/23

Answered by witcher3 last updated on 20/Apr/23

$$\tan \left(\mathrm{a}\right)\tan \left(\mathrm{b}\right)=\frac{\cos (\mathrm{a}-\mathrm{b})-\cos (\mathrm{a}+\mathrm{b})}{\cos (\mathrm{a}-\mathrm{b})+\cos (\mathrm{a}+\mathrm{b})}.$$$$\tan \left(42\right)\tan \left(78\right)=\frac{\cos \left(36\right)+\cos \left(120\right)}{\cos \left(36\right)-\cos \left(120\right)}=\frac{2\cos \left(36\right)-1}{2\cos \left(36\right)+1}$$$$\tan \left(6\right)\tan \left(66\right)=\frac{\cos \left(60\right)-\cos \left(72\right)}{\cos \left(60\right)+\cos \left(72\right)}$$$$=\frac{1-2\cos \left(72\right)}{1+2\cos \left(72\right)}$$$$\iff \frac{1-2\cos \left(72\right)}{1+2\cos \left(72\right)}.\frac{2\cos \left(36\right)-1}{2\cos \left(36\right)+1}=1$$$$\cos \left(36\right)=\mathrm{a}$$$$\iff \frac{1-2({2\mathrm{a}}^2-1)}{1+2({2\mathrm{a}}^2-1)}.\frac{2\mathrm{a}-1}{2\mathrm{a}+1}=1$$$$=\frac{3-{4\mathrm{a}}^2}{{(2\mathrm{a}+1)}^2}=1...\left(\mathrm{E}\right)$$$$\cos (2.36)=\cos (3.36)$$$$\iff {2\mathrm{a}}^2-1={4\mathrm{a}}^3-3\mathrm{a}$$$$\iff {4\mathrm{a}}^3-{2\mathrm{a}}^2-3\mathrm{a}+1=0$$$$\iff (4\mathrm{a}-4)({\mathrm{a}}^2+\frac{\mathrm{a}}{2}-\frac{1}{4})=0$$$$\Rightarrow {4\mathrm{a}}^2+2\mathrm{a}-1=0$$$$3-{4\mathrm{a}}^2=2\mathrm{a}+2$$$${(2\mathrm{a}+1)}^2=1+4\mathrm{a}+{4\mathrm{a}}^2=1+4\mathrm{a}+1-2\mathrm{a}=2+2\mathrm{a}$$$$\left(\mathrm{E}\right)\iff \frac{2\mathrm{a}+2}{2\mathrm{a}+2}=1...\mathrm{True}$$$$$$$$$$$$$$

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