If A= 3^ 9^ + 9^ 27^ + 27^ 81^ and B= 81^ - 3^ , then (B)/(A) is equal to ____ .

Question

If $A=\frac{\sin {3}^{\circ }}{\cos {9}^{\circ }}+\frac{\sin {9}^{\circ }}{\cos {27}^{\circ }}+\frac{\sin {27}^{\circ }}{\cos {81}^{\circ }}$ and $B=	an {81}^{\circ }-	an {3}^{\circ }$, then $\frac{B}{A}$ is equal to

$____$ .

Answer: 2

Accepted Answer

Using $\tan 3\theta - \tan\theta = \dfrac{\sin 2\theta}{\cos 3\theta \cos\theta} = \dfrac{2\sin\theta}{\cos 3\theta}$, i.e. $\dfrac{\sin\theta}{\cos 3\theta} = \dfrac12(\tan 3\theta - \tan\theta)$.

Applying this with $\theta = 3^\circ, 9^\circ, 27^\circ$:

$$\frac{\sin3^\circ}{\cos9^\circ} = \tfrac12(\tan9^\circ-\tan3^\circ), \quad \frac{\sin9^\circ}{\cos27^\circ} = \tfrac12(\tan27^\circ-\tan9^\circ), \quad \frac{\sin27^\circ}{\cos81^\circ} = \tfrac12(\tan81^\circ-\tan27^\circ)$$

Adding, the middle terms telescope:

$$A = \tfrac12(\tan 81^\circ - \tan 3^\circ) = \tfrac12 B$$

Hence $\dfrac{B}{A} = 2$.

Solutions · 1