Find the value of expression $\frac{1}{\cos 290^{\circ}} + \frac{1}{\sqrt3 \sin250^{\circ}}$.
Though this question seems easy, I just seem to get stuck somewhere. This is what I have tried so far: $$\frac{1}{\cos70^{\circ}}-\frac{1}{\sqrt3\sin70^{\circ}}$$ $$=\frac{\sqrt3\sin70^{\circ}-\cos70^{\circ}}{\sqrt3\sin70^{\circ}\cos70^{\circ}}$$
From here, I do not know how to continue. Pls help. Thank you :)
On
The repeated $70°$ as argument of $\cos$ and $\sin$ should remind you of the formulas of the type $\cos(a+b)=...$, $\sin(a+b)=...$.
Then you'll notice that you're missing other $\cos$ and $\sin$ to make use of them. How about multiplying both numerator and denominator by ½ and recognizing some well known values of $\cos$ and $\sin$?
$$\frac{\sqrt{3}\sin 70^\circ-\cos 70^\circ }{\sqrt{3}\sin 70^\circ \cos 70^\circ}=\frac{\frac{\sqrt{3}}{2}\sin 70^\circ-\frac{1}{2}\cos 70^\circ }{\frac{\sqrt{3}}{2}\sin 70^\circ \cos 70^\circ}=$$
$$=\frac{\sin 70^\circ \cos 30^\circ -\sin 30^\circ\cos 70^\circ }{\frac{\sqrt{3}}{4}(2\sin 70^\circ \cos 70^\circ)}=\frac{\sin (70^\circ - 30^\circ)}{\frac{\sqrt{3}}{4}(\sin 140^\circ)}=\frac{\sin 40^\circ}{\frac{\sqrt{3}}{4}(\sin 40^\circ)}=\frac{4}{\sqrt{3}}.$$