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I am quite a baffled now, I am not getting by how it can be written that :

\begin{align*} &\log_{10} \tan 40^\circ \cdot \log_{10} \tan 41^\circ \cdot \log_{10} \tan 42^\circ \cdot\log_{10} \tan 43^\circ \cdots \log_{10} \tan 50^\circ \\ &= \log_{10} \tan 40^\circ + \log_{10} \tan 41^\circ + \log_{10} \tan 42^\circ + \log_{10} \tan 43^\circ + \cdots + \log_{10} \tan 50^\circ \end{align*}

Is it even valid ? If yes,how ?

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    Are you missing `+` signs in the right hand side?2010-11-18
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    @Arturo Magidin :Thanks for the editing.2010-11-18

2 Answers 2

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Yes it is valid.

$\log_{10} \tan 40^{\circ} + \log_{10} \tan 50^{\circ} = \log_{10} \tan 40^{\circ} + \log_{10} \cot 40^{\circ} = \log_{10} 1 = 0$

Similarly combine 41 and 49, 42 and 48 etc.

The product on the left side is $0$ as $\log_{10} \tan 45^{\circ} = 0$.

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    That's awesome, btw what is ur age Moron ? 55+ ? :P2010-11-18
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    @Deb: Check my profile.2010-11-18
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    @Moron:well,I don't remember it to be there last time I checked, ... You are way older than me ;)2010-11-18
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    @Deb: Check again :-)2010-11-18
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    Okay I got it now ... ! :)2010-11-18
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    @Moron:Have you changed your profile since... November?2011-04-06
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    @The Chaz: I probably did.2011-04-06
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HINT: Observe that $$\log_{10} \tan{40} + \log_{10} \tan{50} = \log_{10} \sin{40} - \log_{10} \cos{40} + \log_{10} \sin{50} - \log_{10} \cos{50}$$ Now use the fact that $\sin(90 - x) =\cos{x}$ and see that they cancel out.