$\frac{1}{sin40}-\frac{\sqrt 3}{cos40}$ düzenleyip hertarafı $\frac{1}{2}$ ile çarpalım
$\dfrac{\frac{1}{2}.cos40-\frac{\sqrt 3}{2}.sin40}{\frac{1}{2}sin40.cos40}$
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$\underline{Kural}$;
$2sinx.cosx=sin2x$
$sinx.cosx=\frac{sin2x}{2}$
$sin(30-40)=-sin(40-30)=sin30.cos40-sin40.cos30$
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$sin80=cos10$
$\dfrac{\frac{1}{2}.cos40-\frac{\sqrt 3}{2}.sin40}{\frac{1}{2}sin40.cos40}=\dfrac{-sin10}{\frac{1}{4}.sin80}=-4.tan10$