If \tan ( \pi \over 9 ),x,\tan ( 7\pi \over 18 ) are in arithmetic progression and \tan ( \pi \over 9 ),y,\tan ( 5\pi \over 18 ) are also in arithmetic progression, then |x - 2y| is equal to :

Trigonometric Ratio and Identites PYQ — JEE Main 2021 · 27 Jul · Shift 2 · Q27 with Solution

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Correct answer: C

Use the arithmetic progression condition

If three numbers $a,b,c$ are in A.P., then $b=\frac{a+c}{2}.$

So here, $x=\frac{\tan\left(\frac{\pi}{9}\right)+\tan\left(\frac{7\pi}{18}\right)}{2}$ and $y=\frac{\tan\left(\frac{\pi}{9}\right)+\tan\left(\frac{5\pi}{18}\right)}{2}.$

We need $|x-2y|.$

Rewrite the tangent angles

Note that $\frac{7\pi}{18}=\frac{\pi}{2}-\frac{\pi}{9},$ so $\tan\left(\frac{7\pi}{18}\right)=\tan\left(\frac{\pi}{2}-\frac{\pi}{9}\right)=\cot\left(\frac{\pi}{9}\right).$

Also, $\frac{5\pi}{18}=\frac{\pi}{2}-\frac{2\pi}{9},$ so $\tan\left(\frac{5\pi}{18}\right)=\cot\left(\frac{2\pi}{9}\right).$

Let $t=\tan\left(\frac{\pi}{9}\right)=\tan 20^\circ.$ Then $\tan\left(\frac{7\pi}{18}\right)=\cot 20^\circ=\frac{1}{t}.$

Thus, $x=\frac{t+\frac1t}{2}.$

Also, $y=\frac{\tan 20^\circ+\cot 40^\circ}{2}.$

Use a standard identity

We use $\tan 20^\circ\tan 40^\circ\tan 80^\circ=\sqrt{3}.$

Also, $\tan 80^\circ=\cot 10^\circ.$ A more useful relation here is the standard identity $\tan 20^\circ+\tan 40^\circ+\tan 80^\circ=\tan 20^\circ\tan 40^\circ\tan 80^\circ=\sqrt{3}.$

But a cleaner route is to compute directly using known values:

Since

=\frac{\tan20^\circ+\frac1{\tan20^\circ}}{2},$$ we get $$x=\frac{\sin^2 20^\circ+\cos^2 20^\circ}{2\sin20^\circ\cos20^\circ} =\frac{1}{\sin40^\circ}.$$ So, $$x=\csc40^\circ.$$ Similarly, $$y=\frac{\tan20^\circ+\cot40^\circ}{2} =\frac{\frac{\sin20^\circ}{\cos20^\circ}+\frac{\cos40^\circ}{\sin40^\circ}}{2}.$$ Using $\sin40^\circ=2\sin20^\circ\cos20^\circ$, $$y=\frac{1}{2}\left(\frac{\sin20^\circ}{\cos20^\circ}+\frac{\cos40^\circ}{2\sin20^\circ\cos20^\circ}\right).$$ Putting over common denominator $2\sin20^\circ\cos20^\circ$: $$y=\frac{1}{2}\cdot \frac{2\sin^2 20^\circ+\cos40^\circ}{2\sin20^\circ\cos20^\circ}.$$ Now use $$\cos40^\circ=1-2\sin^2 20^\circ,$$ so numerator becomes $$2\sin^2 20^\circ+(1-2\sin^2 20^\circ)=1.$$ Hence, $$y=\frac{1}{2}\cdot\frac{1}{2\sin20^\circ\cos20^\circ} =\frac{1}{2\sin40^\circ}.$$ Therefore, $$y=\frac{1}{2}\csc40^\circ.$$ 4. **Now compute $x-2y$** Since $$x=\csc40^\circ, \qquad y=\frac12\csc40^\circ,$$ we get $$2y=\csc40^\circ=x.$$ Thus, $$x-2y=0.$$ Hence, $$|x-2y|=0.$$ 5. **Check options** The correct option is $$\boxed{\text{C: }0}.$$

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View written solutionFree

Correct answer: C

Use the arithmetic progression condition

If three numbers $a,b,c$ are in A.P., then $b=\frac{a+c}{2}.$

So here, $x=\frac{\tan\left(\frac{\pi}{9}\right)+\tan\left(\frac{7\pi}{18}\right)}{2}$ and $y=\frac{\tan\left(\frac{\pi}{9}\right)+\tan\left(\frac{5\pi}{18}\right)}{2}.$

We need $|x-2y|.$

Rewrite the tangent angles

Note that $\frac{7\pi}{18}=\frac{\pi}{2}-\frac{\pi}{9},$ so $\tan\left(\frac{7\pi}{18}\right)=\tan\left(\frac{\pi}{2}-\frac{\pi}{9}\right)=\cot\left(\frac{\pi}{9}\right).$

Also, $\frac{5\pi}{18}=\frac{\pi}{2}-\frac{2\pi}{9},$ so $\tan\left(\frac{5\pi}{18}\right)=\cot\left(\frac{2\pi}{9}\right).$

Let $t=\tan\left(\frac{\pi}{9}\right)=\tan 20^\circ.$ Then $\tan\left(\frac{7\pi}{18}\right)=\cot 20^\circ=\frac{1}{t}.$

Thus, $x=\frac{t+\frac1t}{2}.$

Also, $y=\frac{\tan 20^\circ+\cot 40^\circ}{2}.$

Use a standard identity

We use $\tan 20^\circ\tan 40^\circ\tan 80^\circ=\sqrt{3}.$

Also, $\tan 80^\circ=\cot 10^\circ.$ A more useful relation here is the standard identity $\tan 20^\circ+\tan 40^\circ+\tan 80^\circ=\tan 20^\circ\tan 40^\circ\tan 80^\circ=\sqrt{3}.$

But a cleaner route is to compute directly using known values:

Since

=\frac{\tan20^\circ+\frac1{\tan20^\circ}}{2},$$ we get $$x=\frac{\sin^2 20^\circ+\cos^2 20^\circ}{2\sin20^\circ\cos20^\circ} =\frac{1}{\sin40^\circ}.$$ So, $$x=\csc40^\circ.$$ Similarly, $$y=\frac{\tan20^\circ+\cot40^\circ}{2} =\frac{\frac{\sin20^\circ}{\cos20^\circ}+\frac{\cos40^\circ}{\sin40^\circ}}{2}.$$ Using $\sin40^\circ=2\sin20^\circ\cos20^\circ$, $$y=\frac{1}{2}\left(\frac{\sin20^\circ}{\cos20^\circ}+\frac{\cos40^\circ}{2\sin20^\circ\cos20^\circ}\right).$$ Putting over common denominator $2\sin20^\circ\cos20^\circ$: $$y=\frac{1}{2}\cdot \frac{2\sin^2 20^\circ+\cos40^\circ}{2\sin20^\circ\cos20^\circ}.$$ Now use $$\cos40^\circ=1-2\sin^2 20^\circ,$$ so numerator becomes $$2\sin^2 20^\circ+(1-2\sin^2 20^\circ)=1.$$ Hence, $$y=\frac{1}{2}\cdot\frac{1}{2\sin20^\circ\cos20^\circ} =\frac{1}{2\sin40^\circ}.$$ Therefore, $$y=\frac{1}{2}\csc40^\circ.$$ 4. **Now compute $x-2y$** Since $$x=\csc40^\circ, \qquad y=\frac12\csc40^\circ,$$ we get $$2y=\csc40^\circ=x.$$ Thus, $$x-2y=0.$$ Hence, $$|x-2y|=0.$$ 5. **Check options** The correct option is $$\boxed{\text{C: }0}.$$

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