Đáp án:

$3)\\ a)-1\\ b)-1\\ c)1\\ d)=0$

Giải thích các bước giải:

$3)\\ a)\sin^258^\circ+\sin^222^\circ+\sin^232^\circ+\sin^268^\circ-3\\ =\sin^258^\circ+\sin^222^\circ+\cos^2(90^\circ-32^\circ)+\cos^2(90^\circ-68^\circ)-3\\ =\sin^258^\circ+\sin^222^\circ+\cos^258^\circ+\cos^222^\circ-3\\ =\sin^258^\circ+\cos^258^\circ+\sin^222^\circ+\cos^222^\circ-3\\ =1+1-3\\ =-1\\ b)\cot 58^\circ.\cot 32^\circ.\cot 22^\circ.\cot 68^\circ-2\\ =\cot 58^\circ.\tan(90^\circ-32^\circ).\cot 22^\circ.\tan(90^\circ-68^\circ)-2\\ =\cot 58^\circ.\tan 58^\circ.\cot 22^\circ.\tan 22^\circ-2\\ =1.1-2\\ =-1\\ c)\sin^235^\circ+\tan 22^\circ+\sin^255^\circ-\cot 13^\circ:\tan 77 ^\circ-\cot 68^\circ\\ =\sin^235^\circ+\tan 22^\circ+\cos^2(90^\circ-55^\circ)-\cot 13^\circ:\cot(90^\circ- 77 ^\circ)-\tan(90^\circ- 68^\circ)\\ =\sin^235^\circ+\tan 22^\circ+\cos^235^\circ-\cot 13^\circ:\cot 13 ^\circ-\tan 22^\circ\\ =\sin^235^\circ+\cos^235^\circ\\ =1\\ d)\sin 57^\circ-\cos 33^\circ-\cos 64^\circ.\tan 64^\circ+ \sin 64^\circ\\ =\sin 57^\circ-\sin(90^\circ-33^\circ)-\cos 64^\circ.\dfrac{\sin 64^\circ}{\cos 64^\circ}+\sin 64^\circ\\ =\sin 57^\circ-\sin 57^\circ-\sin 64^\circ+\sin 64^\circ\\ =0$