Question

Find of $\cos ec{10}^{\circ }+\cos ec{50}^{\circ }-\cos ec{70}^{\circ }.$

Answer 1

$\csc {10}^{\circ }+\csc {50}^{\circ }-\csc {70}^{\circ }$

$=\frac{1}{\sin {10}^{\circ }}+\frac{1}{\sin (60-10{)}^{\circ }}-\frac{1}{\sin (60+10{)}^{\circ }}$

$=\frac{1}{\sin {10}^{\circ }}+\frac{\sin (60+10{)}^{\circ }-\sin (60-10{)}^{\circ }}{\sin (60+10{)}^{\circ }\sin (60+10{)}^{\circ }}$

$=\frac{1}{\sin {10}^{\circ }}+\frac{2\cos {60}^{\circ }\sin {10}^{\circ }}{\frac{3}{4}{\cos }^2{10}^{\circ }-\frac{1}{4}{\sin }^2{10}^{\circ }}$

$=\frac{1}{\sin {10}^{\circ }}+\frac{4\sin {10}^{\circ }}{3{\cos }^2{10}^{\circ }-{\sin }^2{10}^{\circ }}$

$=\frac{3{\cos }^2{10}^{\circ }-{\sin }^2{10}^{\circ }+4{\sin }^2{10}^{\circ }}{\sin {10}^{\circ }(3-4{\sin }^2{10}^{\circ })}$

$=\frac{3}{\sin {30}^{\circ }}$

$\therefore \csc {10}^{\circ }+\csc {50}^{\circ }-\csc {70}^{\circ }=6$