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Express the number $\sqrt3 \sin(10^\circ) +\dfrac38\tan(10^\circ)$ in the form $\dfrac{\sqrt a} b$, where $a$ and $b$ are integers.

I am sure that trigonometric formulas must be used here, but I cannot see how. I will also be appreciative if someone gives me any hints.

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I think it should $\frac34$ instead of $\frac38$

$$\sin(30^\circ-10^\circ)=\sin20^\circ$$

$$\implies \sin30^\circ\cos10^\circ-\cos30^\circ\sin10^\circ=2\sin10^\circ\cos10^\circ$$

$$\implies \frac12\cos10^\circ-\frac{\sqrt3}2\sin10^\circ=2\sin10^\circ\cos10^\circ$$

$$\implies \frac12-\frac{\sqrt3}2\tan10^\circ=2\sin10^\circ$$ (Dividing by $\cos10^\circ\ne0$)

$$\implies \frac{\sqrt3}2\tan10^\circ+2\sin10^\circ=\frac12$$

Multiply by $\frac{\sqrt3}2$

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  • $\begingroup$ It is 3/8, not 3/4. At first I also thought it was 3/4 and proceeded this way, but it is 3/8. :( $\endgroup$ – user73528 Apr 20 '13 at 15:25
  • $\begingroup$ @efghijkl, I have tried the approach of (math.stackexchange.com/questions/10661/… $a\sin\theta+b\tan\theta=c,$ putting $c=r\sin\alpha,b=r\cos\alpha$ but we need $\alpha=3\cdot10^\circ$ or $180^\circ-10^\circ$, but $\cos\alpha=\frac{\sqrt3}4,$ where $\frac{\sqrt3}2$ would nicely fit $\endgroup$ – lab bhattacharjee Apr 21 '13 at 14:44

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