Find $\frac{\tan 54-\tan 2}{\frac{\sin 2}{\cos 6}+\frac{\sin 6}{\cos 18}+\frac{\sin 18}{\cos 54}}$

Question

Find $$S=\frac{\tan 54-\tan 2}{\frac{\sin 2}{\cos 6}+\frac{\sin 6}{\cos 18}+\frac{\sin 18}{\cos 54}}$$

All the angles are in degrees.

My Try: I tried converting everything to $\sin$ and $\cos$ we get

$$S=\frac{\sin 52}{\cos 2 \cos 54} \times \frac{\cos 6 \cos 18 \cos 54}{\sin 2 \cos 18 \cos 54+\sin 6 \cos 6 \cos 54+\sin 18 \cos 18 \cos 6}$$

any clue here?

Answer 1

From @labbhattacharjee's hint, we have that $$\frac{\sin x}{\cos3x}=\frac{\frac12\sin(3x-x)}{\cos x\cos 3x}=\frac{\sin3x\cos x-\sin x\cos3x}{2\cos x\cos3x}=\frac12\tan3x-\frac12\tan x$$ so $$\require{cancel}\begin{align}\frac{\sin 2}{\cos 6}+\frac{\sin 6}{\cos 18}+\frac{\sin 18}{\cos 54}&=\cancel{\frac12\tan6}-\frac12\tan2+\cancel{\frac12\tan18}-\cancel{\frac12\tan6}+\frac12\tan54-\cancel{\frac12\tan18}\\&=\frac12\left(\tan54-\tan2\right)\end{align}$$ hence $$S=\frac{\tan54-\tan2}{\frac12(\tan54-\tan2)}=\color{red}2$$

Answer 2

Have a look at @labbhattacharjee's comment, we will use that identity in the answer.

\begin{align} \frac{\sin x}{\cos3x} & = \frac{\sin(3x-x)}{2 \cos x \cos 3x} \\ \tag{Angle Difference} & = \frac{\sin 3x \cos x - \cos 3x \sin x}{2\cos x \cos 3x} \\ & = \frac{\cos x \cos 3x(\tan 3x - \tan x)}{2\cos x \cos 3x} \\ &= \frac{\tan 3x - \tan x}{2} \end{align}

We have

\begin{align} S & = \frac{\tan 54-\tan 2}{\frac{\sin 2}{\cos 6}+\frac{\sin 6}{\cos 18}+\frac{\sin 18}{\cos 54}}\\ & = \frac{\tan 54-\tan 2}{{\frac{\tan6- \tan2}{2}}+{\frac{\tan18- \tan6}{2}}+{\frac{\tan54- \tan18}{2}}} \\ & = \frac{ \tan54 - \tan 2}{\frac{\tan 54 - \tan 2}{2}}\\ & = 2 \end{align}