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(1) Evaluation of $\displaystyle \tan \left(\frac{\pi}{7}\right)\cdot \tan \left(\frac{2\pi}{7}\right)\cdot \tan \left(\frac{3\pi}{7}\right) = $

(2) Evaluation of $\displaystyle \tan\left(\frac{\pi}{11}\right)\cdot \tan\left(\frac{2\pi}{11}\right)\cdot \tan\left(\frac{3\pi}{11}\right)\cdot \tan\left(\frac{4\pi}{11}\right)\cdot \tan\left(\frac{5\pi}{11}\right) = $

$\bf{My\; Try::(1):}$ Let $\displaystyle \frac{\pi}{7} = \theta\Rightarrow \pi = 7\theta\Rightarrow (\pi-3\theta) = 4\theta\Rightarrow \tan(\pi-3\theta) = \tan(4\theta)$

So $\displaystyle \tan (3\theta) = -\tan (4\theta)\Rightarrow \frac{3\tan \theta - \tan^3\theta}{1-3\tan^2 \theta} = -\left(\frac{2\tan 2\theta}{1-\tan^2 2\theta}\right) = \frac{4\tan^3 \theta-4\tan \theta}{1+\tan^4 \theta-6\tan^2 \theta}$

So $\displaystyle \left(\frac{3-\tan^2 \theta}{1-3\tan^2 \theta}\right) = \left(\frac{4\tan^2 \theta-4}{1+\tan^4 \theta-6\tan^2 \theta}\right)\Rightarrow \tan^6\theta-21\tan^4 \theta+21\tan^2 \theta-7=0$

Now Let $\displaystyle \tan \theta = y\;,$ then eqn. convert into $\displaystyle y^6-21y^4+21y^2-7=0$

and equation has a roots

$\displaystyle y = \tan \left(\frac{\pi}{7}\right)\;,\tan \left(\frac{2\pi}{7}\right)\;,\tan \left(\frac{3\pi}{7}\right)\;,\tan \left(\frac{4\pi}{7}\right)\;,\tan \left(\frac{5\pi}{7}\right)\;,\tan \left(\frac{6\pi}{7}\right)\;$.

So Product of Roots is

$\displaystyle \tan \left(\frac{\pi}{7}\right)\cdot \tan \left(\frac{2\pi}{7}\right)\cdot \tan \left(\frac{3\pi}{7}\right)\cdot \tan \left(\frac{4\pi}{7}\right)\cdot \tan \left(\frac{5\pi}{7}\right)\cdot \tan \left(\frac{6\pi}{7}\right) = -7$

Now $\displaystyle \tan\left(\frac{4\pi}{7}\right) = \tan\left(\pi-\frac{3\pi}{7}\right)=-\tan\left(\frac{3\pi}{7}\right)$

Similarly $\displaystyle \tan\left(\frac{5\pi}{7}\right) = -\tan\left(\frac{2\pi}{7}\right)$ and $\displaystyle \tan\left(\frac{6\pi}{7}\right) = -\tan\left(\frac{\pi}{7}\right)$

So $\displaystyle \tan\left(\frac{\pi}{7}\right)\cdot \tan\left(\frac{2\pi}{7}\right)\cdot \tan\left(\frac{3\pi}{7}\right)=\sqrt{7}$

I did not understand how can i calculate $(2)$ one using the same method:, bcz it is very lengthy

can we solve the $(1)$ and $(2)$ question any other method.

If yes then plz explain me, Thanks

::

Thanks

Community
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juantheron
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1 Answers1

4

Like Prove the trigonometric identity $(35)$,

$$\prod_{k=1}^m\tan \left(\frac{k \pi}{2m+1}\right)=\sqrt {2m+1}$$ as all the angles lies in $\left(0,\frac\pi2\right)$

lab bhattacharjee
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