Giải thích các bước giải:
1) $\dfrac{sin4x + sin5x + sin6x}{cos4x + cos5x + cos6x}$
$= \dfrac{\left ( sin4x + sin6x \right ) + sin5x}{\left ( cos4x + cos6x \right ) + cos5x}$
$= \dfrac{2sin5xcosx + sin5x}{2cos5xcosx + cos5x}$
$= \dfrac{sin5x\left ( 2cosx + 1 \right )}{cos5x\left ( 2cosx + 1 \right )}$
$= \dfrac{sin5x}{cos5x}$
$= tan5x$
2) $tanx.tan3x = \dfrac{\left ( tan2x + tanx \right )\left ( tan2x - tanx \right )}{\left ( 1 - tanx.tan2x \right )\left ( 1 + tanx.tanx \right )} = \dfrac{tan^{2}2x - tan^{2}x}{1 - tan^{2}2x.tan^{2}x}$
