Question

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

• A. 0
• B. 1
• C. 3
• D. 4

Hint:

When dealing with sums of tangent functions, look for special angle relationships. If angles sum to $90^\circ$ ($A+B=\pi/2$), then $\tan A = \cot B$. Also, the identity $\tan A + \tan B = \frac{\sin(A+B)}{\cos A \cos B}$ is very useful.

Answer 1

The Correct Option is A

For an arithmetic progression, \[ 2x = \tan\!\left(\frac{\pi}{9}\right) + \tan\!\left(\frac{7\pi}{18}\right) \] Since \[ \frac{7\pi}{18} = \frac{\pi}{2} - \frac{\pi}{9}, \quad \tan\!\left(\frac{7\pi}{18}\right)=\cot\!\left(\frac{\pi}{9}\right) \] \[ 2x = \tan A + \cot A = \frac{1}{\sin A \cos A} = \frac{2}{\sin(2A)} \] with $A=\frac{\pi}{9}$, \[ x = \frac{1}{\sin(\frac{2\pi}{9})} \] Now, \[ 2y = \tan\!\left(\frac{\pi}{9}\right) + \tan\!\left(\frac{5\pi}{18}\right) \] Using identity: \[ \tan A + \tan B = \frac{\sin(A+B)}{\cos A \cos B} \] Here, \[ A+B = \frac{\pi}{9} + \frac{5\pi}{18} = \frac{7\pi}{18} \] \[ 2y = \frac{\sin(\frac{7\pi}{18})}{\cos(\frac{\pi}{9})\cos(\frac{5\pi}{18})} \] Since $\sin(\frac{7\pi}{18})=\cos(\frac{\pi}{9})$, \[ 2y = \frac{1}{\cos(\frac{5\pi}{18})} = \frac{1}{\sin(\frac{2\pi}{9})} \] Thus, \[ x = 2y \Rightarrow |x-2y|=0 \]