Question

$\frac{2\cos {75}^{\circ }}{\sin {15}^{\circ }}-\frac{\sin {35}^{\circ }}{2\cos {55}^{\circ }}$ is $1\frac{1}{m}$. m is

Answer 1

$\frac{2\cos {75}^{\circ }}{\sin {15}^{\circ }}-\frac{\sin {35}^{\circ }}{2\cos {55}^{\circ }}$

= $\frac{2\cos (90-15{)}^{\circ }}{\sin {15}^{\circ }}-\frac{\sin (90-55{)}^{\circ }}{2\cos {55}^{\circ }}$

= $\frac{2\sin {15}^{\circ }}{\sin {15}^{\circ }}-\frac{\cos {55}^{\circ }}{2\cos {55}^{\circ }}$

= $2-\frac{1}{2}$

= $\frac{3}{2}$

= $1\frac{1}{2}$