3
$\begingroup$

To prove:

$$\frac{1}{\sin 10^\circ}-\frac{\sqrt 3}{\cos 10^\circ}=4$$

I tried taking lcm but could not get to anything.

  • 0
    are the angles in degrees? – 2017-02-11
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    @Dr.SonnhardGraubner yes edited – 2017-02-11
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    Hint: write it as $\;\frac{1}{2} \cos 10^\circ - \frac{\sqrt{3}}{2} \sin 10^\circ = \sin 20^\circ\;$ and remember what $\sin 30^\circ, \cos 30^\circ$ are. – 2017-02-11
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    @dxiv oh thanks i solved it – 2017-02-11

3 Answers 3

8

\begin{align*} \frac {1}{\sin 10^\circ} - \frac {\sqrt 3}{\cos 10^\circ} &= \frac {\cos 10^\circ - \sqrt 3\sin 10^\circ}{\sin 10^\circ \cos 10^\circ} \\ &= 2\left(\frac {\frac 12\cos 10^\circ - \frac {\sqrt 3}{2}\sin 10^\circ}{\sin 10^\circ \cos 10^\circ}\right) \\ &= 2\left(\frac {\sin 30^\circ\cos 10^\circ - \cos 30^\circ\sin 10^\circ}{\sin 10^\circ \cos 10^\circ}\right) \\ (a) \ldots &= 2\left(\frac {\sin (30^\circ-10^\circ)}{\sin 10^\circ \cos 10^\circ}\right) \\ &= 2\left(\frac {\sin 20^\circ}{\sin 10^\circ \cos 10^\circ}\right) \\ (b) \ldots &= 2\left(\frac {2\sin 10^\circ \cos 10^\circ}{\sin 10^\circ \cos 10^\circ}\right) \\ &= 4 \end{align*}

$(a) \ldots \rightarrow \sin (X-Y)=\sin X \cos Y - \sin Y \cos X $

$(b) \ldots \rightarrow \sin (2X)=2\sin X \cos X $

1

Set $$ x=\dfrac{1}{\sin10^\circ}-\dfrac{\sqrt{3}}{\cos10^\circ},\quad y=\sin10^\circ\cos10^\circ=\dfrac12\sin20^\circ \ne 0. $$ We have \begin{eqnarray} \dfrac12xy&=&\dfrac12\cos10^\circ-\dfrac{\sqrt{3}}{2}\sin10^\circ=\sin30^\circ\cos10^\circ-\cos30^\circ\sin10^\circ=\sin(30^\circ-10^\circ)=\sin20^\circ, \end{eqnarray} i.e. $$ \dfrac12xy=2y. $$ Since $y\ne 0$, it follows that $$ x=4. $$

1

$$ \frac{\sin 20}{\sin20}=1$$

$$\frac {\sin(30-10)}{\frac12\sin 10 \cos 10}=1 $$ and hence after expanding $\sin(30-10)$ we can have the result.

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    how clever ! how did you think that out – 2017-02-11