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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 :)

2 Answers2

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$$\frac{\sqrt{3}\sin 70^\circ-\cos 70^\circ }{\sqrt{3}\sin 70^\circ \cos 70^\circ}=\frac{\frac{\sqrt{3}}{2}\sin 70^\circ-\frac{1}{2}\cos 70^\circ }{\frac{\sqrt{3}}{2}\sin 70^\circ \cos 70^\circ}=$$

$$=\frac{\sin 70^\circ \cos 30^\circ -\sin 30^\circ\cos 70^\circ }{\frac{\sqrt{3}}{4}(2\sin 70^\circ \cos 70^\circ)}=\frac{\sin (70^\circ - 30^\circ)}{\frac{\sqrt{3}}{4}(\sin 140^\circ)}=\frac{\sin 40^\circ}{\frac{\sqrt{3}}{4}(\sin 40^\circ)}=\frac{4}{\sqrt{3}}.$$

Жека
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  • It may contain mistakes, since a lot of time has pass since I did any trigonometry. – Жека Feb 14 '19 at 12:57
  • Why not leave the OP solve the problem with the indications...? – Joce Feb 14 '19 at 13:22
  • I am new to this site, it is customary ? – Жека Feb 14 '19 at 21:42
  • Well, I had given in my answer below the route towards your answer. As this is obviously homework, writing out the answer from this is not beneficial. It is true that this SE (compared to physics.SE e.g.) is happy to provide answers rather than guidance for homework. – Joce Feb 15 '19 at 08:23
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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 $\sin$?

Joce
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