Giải giúp e 2 bài này vs ạ em đang cần gấp. Em cảm ơn trc ạ

Đáp án + Giải thích các bước giải:

a) $A = \dfrac{\sin 30^o \cos 60^o}{\tan 45^o}$
$= \dfrac{\sin 30^o \cos (90^o - 30^o)}{1}$
$= \dfrac{\sin 30^o \sin 30^o}{1}$

$= \sin^2 30^o$

$= \bigg(\dfrac{1}{2}\bigg)^2 = \dfrac{1}{4}$

b) $B = \dfrac{\sin 60^o + \cos 30^o}{\tan 30^o \cot 60^o}$
$= \dfrac{\sin 60^o + \cos (90^o - 60^o)}{\tan 30^o \cot (90^o - 30^o)}$

$= \dfrac{\sin 60^o + \sin 60^o}{\tan 30^o \tan 30^o}$
$= \dfrac{2\sin 60^o}{\tan^2 30^o}$

$= \dfrac{2 . \frac{\sqrt{3}}{2}}{\Big(\frac{\sqrt{3}}{3}\Big)^2}$

$= \dfrac{\sqrt{3}}{\frac{1}{3}}$
$= 3\sqrt{3}$

c) $C = \sin^2 45^o + \cos^2 30^o - \tan^2 60^o$

$= \bigg(\dfrac{\sqrt{2}}{2}\bigg)^2 + \bigg(\dfrac{\sqrt{3}}{2}\bigg)^2 - (\sqrt{3})^2$

$= \dfrac{1}{2} + \dfrac{3}{4} - 3$

$= -\dfrac{7}{4}$

Bài 4:

a) $A = \dfrac{\tan 25^o}{\cot 65^o}$

$= \dfrac{\tan 25^o}{\cot (90^o - 25^o)}$
$= \dfrac{\tan 25^o}{\tan 25^o}$

$= 1$

b) $B = \sin 25^o + \cos 25^o - \sin 65^o - \cos 65^o$

$= \sin 25^o + \cos 25^o - \sin (90^o - 25^o) - \cos (90^o - 25^o)$
$= \sin 25^o + \cos 25^o - \cos 25^o - \sin 25^o$

$= 0$

c) $C = \tan 44^o + \sin 17^o - \cot 46^o - \cos 73^o$

$= \tan 44^o + \sin 17^o - \cot (90^o - 44^o) - \cos (90^o - 17^o)$
$= \tan 44^o + \sin 17^o - \tan 44^o - \sin 17^o$
$= 0$

d) $D = \tan 1^o \tan 15^o \tan 75^o \tan 89^o$

$= \tan 1^o \tan 89^o \tan 15^o \tan 75^o$

$= \tan 1^o \tan (90^o - 1^o) \tan 15^o \tan (90^o - 15^o)$
$= \tan 1^o \cot 1^o \tan 15^o \cot 15^o$

$= 1 . 1$

$= 1$

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