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Trigonometric Equations

Section: Mathematics

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Question : 2 of 24

Marks: +1, -0

\tan \frac{2\pi}{7} \cdot \tan \frac{4\pi}{7}+\tan \frac{4\pi}{7} \cdot \tan \frac{\pi}{7}+\tan \frac{\pi}{7} \cdot \tan \frac{2\pi}{7}=

[AP EAPCET 24 May 2025 S1]

$7$
$-7$
$3$
$-3$

Solution:

Given, expression is

\tan \left(\frac{2\pi}{7}\right) \cdot \tan \left(\frac{4\pi}{7}\right)+\tan \left(\frac{4\pi}{7}\right) \cdot \tan \left(\frac{\pi}{7}\right)+\tan \left(\frac{\pi}{7}\right) \cdot \tan \left(\frac{2\pi}{7}\right)

Let

x=\frac{\pi}{7}, y=\frac{2\pi}{7}

and

z=\frac{4\pi}{7}

So, the expression becomes

\tan(y) \cdot \tan(z)+\tan(z) \cdot \tan(x)+\tan(x) \cdot \tan(y)

Now,

x+y+z=\frac{\pi}{7}+\frac{2\pi}{7}+\frac{4\pi}{7}=\frac{7\pi}{7}=\pi

Since,

x+y+z=\pi

, we can use the identify

\tan(x) \cdot \tan(y)+\tan(y) \cdot \tan(z)+\tan(z) \cdot \tan(x)=-7

\therefore \tan \left(\frac{\pi}{7}\right) \tan \left(\frac{2\pi}{7}\right)+\tan \left(\frac{4\pi}{7}\right) \cdot \tan \left(\frac{2\pi}{7}\right)+\tan \left(\frac{4\pi}{7}\right) \tan \left(\frac{\pi}{7}\right)=-7

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