Đáp án + Giải thích các bước giải:

$E = 4 \tan 32^o . \cos 60^o . \cot 148^o + \dfrac{5 \cot^2 108^o}{1 + \tan^2 18^o} + 5 \sin^2 72^o$

$= 4 \tan 32^o . \dfrac{1}{2} . \dfrac{1}{\tan 148^o} + \dfrac{5 \cot^2 (180^o - 72^o)}{\frac{1}{\cos^2 18^o}} + 5 \sin^2 72^o$

$= 2 \tan 32^o . \dfrac{1}{\tan (180^o - 32^o)} + 5(-\cot 72^o)^2 . \cos^2 18^o + 5 \sin^2 72^o$

$= 2 \tan 32^o . \bigg(-\dfrac{1}{\tan 32^o}\bigg) + 5[-\cot (90^o - 18^o)]^2 . \cos^2 18^o + 5\sin^2 72^o$

$= -2 + 5[-\tan 18^o]^2 . \cos^2 18^o + 5\sin^2 72^o$

$= -2 + 5 \tan^2 18^o . \cos^2 18^o + 5\sin^2 72^o$

$= -2 + \dfrac{5\sin^2 18^o}{\cos^2 18^o} . \cos^2 18^o + 5\sin^2 72^o$

$= -2 + 5\sin^2 18^o + 5\sin^2 72^o$

$= -2 + 5(\sin^2 18^o + \sin^2 72^o)$

$= -2 + 5[\sin^2 18^o + \sin (90^o - 18^o)]^2$

$= -2 + 5(\sin^2 18^o + \cos^2 18^o)$

$= -2 + 5$

$= 3$