Question

Sin160°cos110°+sin290°cos340°+tg110°tg160°

Answer 1

`\sin 160^(\circ) \cos 110^(\circ) + \sin 290^(\circ) \cos 340^(\circ) + \tan 110^(\circ) \tan 160^(\circ)\\\\=\sin\left(2 * 90^(\circ) - 20^(\circ) \right) \cos \left(90^(\circ) + 20^(\circ) \right) + \sin \left(3 * 90^(\circ)+20^(\circ)\right) \cos\left(4 * 90^(\circ)-20^{\circ \right) + 20^(\circ) \right) + \tan \left( 90^(\circ) +20^(\circ)\right) \tan\left( 2 * 90^(\circ) - 20^(\circ) \right)\\\\`

`=-\sin 20^(\circ) \sin 20^(\circ) -\cos 20^(\circ) \cos 20^(\circ)+\cot 20^(\circ) \tan 20^(\circ)\\\\=-\left(\sin^2 20^(\circ) + \cos^2 20^(\circ) \right) + \tan 20^(\circ) \cdot \frac 1{\tan 20^(\circ)}\\\\=-1 + 1\\\\=0`

Answer 2

-0.0092

Explanation:

sin160°cos110° + sin290°cos340° + tan110°tan160°

(0.34)(-0.34) + (-0.94)(0.94) + (-2.75)(-0.36)

- 0.1156 - 0.8836 + 0.99

-0.0092

I hope this helps!