Question

Find the value of: $\frac{\sin {41}^{\circ }}{2\cos {49}^{\circ }}-\frac{2\sec {32}^{\circ }}{3\csc {58}^{\circ }}$

Answer 1

$\frac{\sin {41}^{\circ }}{2\cos {49}^{\circ }}-\frac{2\text{sec}{32}^{\circ }}{3\text{cosec}{58}^{\circ }}=\frac{\sin ({90}^{\circ }-{49}^{\circ })}{2\cos {49}^{\circ }}-\frac{2\text{sec}({90}^{\circ }-{58}^{\circ })}{3\text{cosec}{58}^{\circ }}=\frac{\cos {49}^{\circ }}{2\cos {49}^{\circ }}-\frac{2\text{cosec}{58}^{\circ }}{3\text{cosec}{58}^{\circ }}=\frac{1}{2}-\frac{2}{3}=\frac{3-4}{6}=-\frac{1}{6}$