Đáp án:
Bài 1:
$a)$
Ta có: $\tan37^o=\cot(90^o-37^o)=\cot53^o$
Mà $\cot53^o>\cot41^o$
Nên $\tan37^o>\cot41^o$
$b)$
$A=\tan60^o.\sin^240^o+\cot30^o.\sin^250^o$
$=\tan60^o.\sin^240^o+\tan(90^o-30^o).\cos^2(90^o-50^o)$
$=\tan60^o.\sin^240^o+\tan60^o.\cos^240^o$
$=\tan60^o.(\sin^240^o+\cos^240^o)$
$=\tan60^o$
$=\sqrt{3}$
$c)$
Ta có:
$+)\sin^2a+\cos^2a=1$
$⇔(\dfrac{1}{2})^2+\cos^2a=1$
$⇔\cos^2a=\dfrac{3}{4}$
$⇔\cos a=\dfrac{\sqrt{3} }{2}$
$+)\tan a =\dfrac{\sin a}{\cos a}$
$\tan a=\dfrac{1 }{\sqrt{3}}$
$+)\cot a=\dfrac{1 }{\tan a}$
$⇔\cot a=\sqrt{3}$
