Đáp án:
$B=0$
Giải thích các bước giải:
$B=tan($ $\dfrac{x}{2}$ $)$ $-$ $\dfrac{sin(x)}{1+cos(x)}$
$B=tan($ $\dfrac{x}{2}$ $)$ $-$ $\dfrac{2sin(\dfrac{x}{2}) cos(\dfrac{x}{2}) }{1+2cos^2(\dfrac{x}{2})-1}$
$B=tan($ $\dfrac{x}{2}$ $)$ $-$ $\dfrac{2sin(\dfrac{x}{2}) cos(\dfrac{x}{2}) }{2cos^2(\dfrac{x}{2})}$
$B=tan($ $\dfrac{x}{2}$ $)$ $-$ $\dfrac{sin(\dfrac{x}{2}) } {cos(\dfrac{x}{2})}$
$B=tan($ $\dfrac{x}{2}$ $)$ $-$ $tan($$\dfrac{x}{2}$ $)$ $=0$
