Này thì khó :))Câu a.
$B=tana+\frac{tana+\sqrt{3}}{1-\sqrt{3}tana}+\frac{tana-\sqrt{3}}{1+\sqrt{3}tana}$
$=\frac{tana(1-3tan^2a)+(tana+\sqrt{3})(1+\sqrt{3}tana)+(tana-\sqrt{3})(1-\sqrt{3}tana)}{1-3tan^2a}$
$=\frac{9tana-3tan^3a}{1-3tan^2a}=\frac{3sina.cosa(3cos^2a-sin^2a)}{cos^2a(cos^2a-3sin^2a)}$
$=\frac{3sina[3(1-sin^2a)-sin^2a]}{cosa[cos^2a-3(1-cos^2a)]}=\frac{3sina(3-4sin^2a)}{cosa(4cos^2a-3)}$
$=\frac{3(3sina-4sin^3a)}{4cos^3a-3cosa}=3.\frac{sin3a}{cos3a}=3tan3a$