JEE MAIN - Mathematics (2021 - 27th July Evening Shift - No. 5)
If $$\tan \left( {{\pi \over 9}} \right),x,\tan \left( {{{7\pi } \over {18}}} \right)$$ are in arithmetic progression and $$\tan \left( {{\pi \over 9}} \right),y,\tan \left( {{{5\pi } \over {18}}} \right)$$ are also in arithmetic progression, then $$|x - 2y|$$ is equal to :
4
3
0
1
Explanation
$$x = {1 \over 2}\left( {\tan {\pi \over 9} + \tan {{7\pi } \over {18}}} \right)$$
and $$2y = \tan {\pi \over 9} + \tan {{5\pi } \over {18}}$$
If we interpret the angles in degrees (as suggested by the numbers 20, 50, and 70), we have :
$$x = \frac{1}{2} \left( \tan 20^\circ + \tan 70^\circ \right),$$
and
$$2y = \tan 20^\circ + \tan 50^\circ.$$
The expression for $|x - 2y|$ is then :
$$|x - 2y| = \left|\frac{\tan 20^\circ + \tan 70^\circ}{2} - \left( \tan 20^\circ + \tan 50^\circ \right)\right|.$$
$$|x - 2y| = \left|\frac{\tan 20^\circ + \tan 70^\circ - 2 \tan 20^\circ - 2 \tan 50^\circ}{2}\right|,$$
which simplifies to :
$$|x - 2y| = \left|\frac{\tan 70^\circ - \tan 20^\circ - 2 \tan 50^\circ}{2}\right|.$$
We know,
$$ \begin{aligned} & \tan 70=\frac{\tan 20+\tan 50}{1-\tan 20 \tan 50} \\\\ &\Rightarrow \tan 70-\tan 70 \cdot \tan 20 \tan 50=\tan 20+\tan 50 \\\\ &\Rightarrow \tan 70-\tan 50-\tan 20-\tan 50=0 \\\\ &\Rightarrow \tan 70-\tan 20-2 \tan 50=0 \end{aligned} $$
$$ \therefore $$ $$|x - 2y| = \left|\frac{\tan 70^\circ - \tan 20^\circ - 2 \tan 50^\circ}{2}\right|$$ = $$\left| {{0 \over 2}} \right|$$ = 0
and $$2y = \tan {\pi \over 9} + \tan {{5\pi } \over {18}}$$
If we interpret the angles in degrees (as suggested by the numbers 20, 50, and 70), we have :
$$x = \frac{1}{2} \left( \tan 20^\circ + \tan 70^\circ \right),$$
and
$$2y = \tan 20^\circ + \tan 50^\circ.$$
The expression for $|x - 2y|$ is then :
$$|x - 2y| = \left|\frac{\tan 20^\circ + \tan 70^\circ}{2} - \left( \tan 20^\circ + \tan 50^\circ \right)\right|.$$
$$|x - 2y| = \left|\frac{\tan 20^\circ + \tan 70^\circ - 2 \tan 20^\circ - 2 \tan 50^\circ}{2}\right|,$$
which simplifies to :
$$|x - 2y| = \left|\frac{\tan 70^\circ - \tan 20^\circ - 2 \tan 50^\circ}{2}\right|.$$
We know,
$$ \begin{aligned} & \tan 70=\frac{\tan 20+\tan 50}{1-\tan 20 \tan 50} \\\\ &\Rightarrow \tan 70-\tan 70 \cdot \tan 20 \tan 50=\tan 20+\tan 50 \\\\ &\Rightarrow \tan 70-\tan 50-\tan 20-\tan 50=0 \\\\ &\Rightarrow \tan 70-\tan 20-2 \tan 50=0 \end{aligned} $$
$$ \therefore $$ $$|x - 2y| = \left|\frac{\tan 70^\circ - \tan 20^\circ - 2 \tan 50^\circ}{2}\right|$$ = $$\left| {{0 \over 2}} \right|$$ = 0
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