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Mathematics Question on Trigonometric Equations

sin⁡70∘ +cos⁡40∘cos⁡70∘+sin⁡40∘\frac {\sin 70^\circ\, +\cos 40^\circ} {\cos 70^\circ + \sin 40^\circ}=

A

13\frac {1}{\sqrt {3}}

B

3\sqrt {3}

C

12\frac {1}{2}

D

11

Answer

3\sqrt {3}

Explanation

Solution

sin⁡70∘ +cos⁡40∘cos⁡70∘+sin⁡40∘\frac {\sin 70^\circ\, +\cos 40^\circ} {\cos 70^\circ + \sin 40^\circ} =sin (90∘−20∘)+sin (90∘−40∘)cos (90∘−20∘)+sin (90∘−50∘)=\frac{sin\, \left(90^{\circ}-20^{\circ}\right)+sin\,\left(90^{\circ}-40^{\circ}\right)}{cos\,\left(90^{\circ}-20^{\circ}\right)+sin\,\left(90^{\circ}-50^{\circ}\right)} =cos 20∘+cos 40∘cos 70∘+cos 50∘=\frac{cos\,20^{\circ}+cos\,40^{\circ}}{cos\,70^{\circ}+cos\,50^{\circ}} =2 cos 30∘ cos 10∘2 cos 60∘ cos 10∘=2 cos 30∘2 cos 60∘=\frac{2\,cos\,30^{\circ}\,cos\,10^{\circ}}{2\,cos\,60^{\circ}\,cos\,10^{\circ}}=\frac{2\,cos\,30^{\circ}}{2\,cos\,60^{\circ}} =2(3/2)2(12)=3=\frac{2\left(\sqrt{3}/2\right)}{2\left(\frac{1}{2}\right)}=\sqrt{3}