Sin(-810)+cos(-900)+tg(-395)ctg575

Решите задачу:

$\sin(-810^\circ)+\cos(-900^\circ)+\tan(-395^\circ)\cot575^\circ=\\=-\sin(810^\circ)+\cos(900^\circ)-\tan(395^\circ)\cot575^\circ=\\=-\sin(2\cdot360^\circ+90^\circ)+\cos(2\cdot360^\circ+180^\circ)-\\-\tan(360^\circ+35^\circ)\cot(360^\circ+215^\circ)=\\=-\sin(90^\circ)+\cos(180^\circ)-\tan(35^\circ)\cot(215^\circ)=\\=-2-\tan(35^\circ)\cot(180^\circ+35^\circ)=\\=-2-\tan(35^\circ)\cot(35^\circ)=-2-1=-3$