Question

The value of $\text{cosec}{10}^{\circ }+\text{cosec}{50}^{\circ }-\text{cosec}{70}^{\circ }$ is

Answer 1

$cosec{10}^{\circ }+cosec{50}^{\circ }-cosec{70}^{\circ }$
$=\frac{1}{\sin {10}^{\circ }}+\frac{1}{\sin {50}^{\circ }}-\frac{1}{\sin {70}^{\circ }}$
$=\frac{2\sin {50}^{\circ }\sin {70}^{\circ }+2\sin {10}^{\circ }\sin {70}^{\circ }-2\sin {10}^{\circ }\sin {50}^{\circ }}{2\sin {10}^{\circ }\sin {50}^{\circ }\sin {70}^{\circ }}$
$=\frac{2\sin {70}^{\circ }(\sin {50}^{\circ }+\sin {10}^{\circ })-2\sin {10}^{\circ }\sin {50}^{\circ }}{2\sin {10}^{\circ }\sin ({60}^{\circ }-{10}^{\circ })\sin ({60}^{\circ }+{10}^{\circ })}$
$=\frac{2\sin {70}^{\circ }(2\sin {30}^{\circ }\cos {20}^{\circ })-\cos {40}^{\circ }+\cos {60}^{\circ }}{2\sin {10}^{\circ }\sin ({60}^{\circ }-{10}^{\circ })\sin ({60}^{\circ }+{10}^{\circ })}$
$=\frac{2{\sin }^2{70}^{\circ }-\cos {40}^{\circ }+\frac{1}{2}}{2\sin {10}^{\circ }({\sin }^2{60}^{\circ }-{\sin }^2{10}^{\circ })}$
$=\frac{1-\cos {140}^{\circ }-\cos {40}^{\circ }+\frac{1}{2}}{2\left(\frac{3\sin {10}^{\circ }-4{\sin }^3{10}^{\circ }}{4}\right)}$
$=\frac{\cos {40}^{\circ }-\cos {40}^{\circ }+\frac{3}{2}}{\frac{\sin {30}^{\circ }}{2}}$
$=\frac{\frac{3}{2}}{\frac{1}{4}}=6$