Question

Without using trigonometric tables evaluate :

$\frac{\sin {63}^{\circ }}{\cos {27}^{\circ }}+\frac{\cos {39}^{\circ }}{\sin {51}^{\circ }}-\sin {23}^{\circ }\cdot \sec {67}^{\circ }+{\csc }^2{30}^{\circ }$

Answer 1

$\frac{\sin {63}^{\circ }}{\cos {27}^{\circ }}+\frac{\cos {39}^{\circ }}{\sin {51}^{\circ }}-\sin {23}^{\circ }\cdot \sec {67}^{\circ }+{\csc }^2{30}^{\circ }$

$=\frac{\cos ({90}^{\circ }-{65}^{\circ })}{\cos {25}^{\circ }}+\frac{\sin ({90}^{\circ }-{32}^{\circ })}{\sin {58}^{\circ }}-\cos ({90}^{\circ }-{28}^{\circ })\times \frac{1}{\cos {62}^{\circ }}+{\left(2\right)}^2$

$=\frac{\cos {25}^{\circ }}{\cos {25}^{\circ }}+\frac{\sin {58}^{\circ }}{\sin {58}^{\circ }}-\cos {62}^{\circ }\times \frac{1}{\cos {62}^{\circ }}+4$

$=1+1-1+4$

$=5$