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If $x=\cot6^\circ\cot42^\circ$ and $y=\tan66^\circ\tan78^\circ$, then

A) $2x=y$

B) $x=2y$

C) $x=y$

D) $2x=3y$

I seriously don’t know where to start. I don’t need the complete answer, but a starting statement which would give me the direction to solve it would be helpful.

Thanks a lot.

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Aditya
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  • $\tan 78^\circ =\cot 12^\circ, \cot 42^\circ=\tan 48^\circ$. Now you have some double angles around. No guarantees. – Ross Millikan Aug 10 '19 at 15:21
  • You commented on @RossMillikan's now-deleted answer (repurposed into the comment above) that you "already tried that". Please ... Always try to include as much as possible of the attempts you've made (and don't write "I seriously don't know where to start" when you have made some attempts), so that people don't waste their time telling you things you already know. – Blue Aug 10 '19 at 16:18
  • Will keep that in mind. I was just saying that what Ross told me was the most basic step, and anyone would have tried that. So I figured it was obvious and said I needed more information than that. – Aditya Aug 10 '19 at 16:43
  • https://math.stackexchange.com/questions/455070/proving-a-fact-tan6-circ-tan42-circ-tan12-circ-tan24-cir – lab bhattacharjee Aug 12 '19 at 05:12

1 Answers1

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as @Ross said $x = \cot 6 \tan 48 , y = \cot 24 \cot 12$ then we look at

$$\frac{x}{y} = \dfrac{\cot 6 \tan 48 }{\cot 24 \cot 12} = \frac{\cot 6 \tan 24}{\cot 48 \cot 12}$$

$$\cot 6 \tan 24 = \frac{\cos 6 \sin 24}{\sin 6 \cos 24}= \frac{\sin 30 + \sin 18}{\sin 30 - \sin 18} = \frac{1 + 2\sin 18 }{1 - 2\sin 18} $$

I used $$\sin a \cos b = 0.5 ( \sin (a+b) + \sin (a-b)) \\ \cos a \sin b = 0.5 ( \sin (a+b) - \sin (a-b))$$

You need also

$$\cos a \cos b = 0.5 ( \cos (a+b) + \cos (a-b) ) \\ \sin a \sin b = 0.5 ( \cos (a-b) - \cos (a+b))$$ to simplify the Denominator can you proceed ?

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