The value of $$\large \displaystyle e^{\log(\tan 1^\circ) + \log(\tan 2^\circ)+ \cdots+\log(\tan 89^\circ)}$$ Base is $10$. I guess it should simplify to $\large\displaystyle e^{89 \log(\tan 1^\circ)}$. Please help. I tried attempting the problem. The options in book are

$A.\ 0$

$B.\ e$

$C.\ \dfrac1e$

$D.\ \text{None of these}$

The answer is none of these. That's why I am confused.

  • $\begingroup$ There is a ......... In place of till. $\endgroup$
    – geek101
    Jul 29, 2014 at 9:24

1 Answer 1



Use $\displaystyle\log a+\log b=\log(ab)$ where all the logarithms remain defined

and $$\tan(90^\circ-x)=\cot x=\frac1{\tan x}=(\tan x)^{-1}$$

$$\implies \log[\tan(90^\circ-x)]=?$$

See also: $ \tan 1^\circ \cdot \tan 2^\circ \cdot \tan 3^\circ \cdots \tan 89^\circ$

  • $\begingroup$ Ahh I got it. T $\endgroup$
    – geek101
    Jul 29, 2014 at 9:30
  • $\begingroup$ So the answer is 1? $\endgroup$
    – geek101
    Jul 29, 2014 at 9:30
  • $\begingroup$ @user166748, $$e^{\log 1}=e^0=?$$ $\endgroup$ Jul 29, 2014 at 9:31
  • $\begingroup$ Yaa that's what I calculated. Thanks a lot sir. $\endgroup$
    – geek101
    Jul 29, 2014 at 9:32
  • $\begingroup$ @5xum what do you mean? I know that this exactly what I needed so I accepted it. $\endgroup$
    – geek101
    Jul 29, 2014 at 11:59

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