Đáp án:
\(A = \tan3x\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad\ A = \dfrac{\sin x + \sin3x + \sin5x}{\cos x + \cos3x + \cos5x}\\
\to A = \dfrac{(\sin x + \sin5x) + \sin3x}{(\cos x + \cos5x) + \cos3x}\\
\to A = \dfrac{2\sin3x\cos2x + \sin3x}{2\cos3x\cos2x + \cos3x}\\
\to A = \dfrac{\sin3x(2\cos2x + 1)}{\cos3x(2\cos2x + 1)}\\
\to A = \dfrac{\sin3x}{\cos3x}\\
\to A = \tan3x
\end{array}\)