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Question:

If $\tan\frac{\pi}{9},x,\tan\frac{5\pi}{18}$ are in arithmetic progression and $\tan\frac{\pi}{9},y,\tan\frac{7\pi}{18}$ are, as well, then

$(A)\;x=y\qquad(B)\;x=2y\qquad(C)\;2x=y\qquad(D)\;x+y=\frac{\pi}{2}$

My Attempt:

I used $2b=a+c$ ($\because$ It is an A.P.) for both cases and the subtracted both equations to get an equation in $x$ and $y$ on LHS and tan terms on RHS.

I then converted $\tan$ to $\sin$ and $\cos$ ($\tan A-\tan B=\sin(A-B)/\cos A\cos B$) and then converted the $\cos$ terms in denominator to $\cos(A+B)+\cos(A-B)$.

But I cannot find what to do further.

Kindly help.

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1 Answer

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I shall convert everything to degrees, for convenience.

Note that $x = \frac 12(\tan 50 + \tan 20) = \frac 12(\tan 70(1-\tan 50 \tan 20) )= \frac 12(\tan 70 - \tan 50)$, since $\tan 70 \tan 20 = 1$.

Adding the previous two expressions, $4x = (\tan 70 + \tan 50 -\tan 50 + \tan 20)= \tan 70 + \tan 20$. We already know that $2y = \tan 70 + \tan 20$.

It is easy to see that $2x=y$ from this.

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