Question

Evaluate
$\left(i\right)\frac{\sin {18}^{\circ }}{\cos {72}^{\circ }}$
$\left(ii\right)\cos {48}^{\circ }-\sin {42}^{\circ }$

• A. $\left(i\right)0\left(ii\right)0$
• B. $\left(i\right)0\left(ii\right)1$
• C. $\left(i\right)1\left(ii\right)0$
• D. Data insufficient
  

Answer 1

The correct option is C $\left(i\right)1\left(ii\right)0$

$\left(i\right)\frac{\sin {18}^{\circ }}{\cos {72}^{\circ }}=\frac{\sin {18}^{\circ }}{\cos ({90}^{\circ }-{18}^{\circ })}=\frac{\sin {18}^{\circ }}{\sin {18}^{\circ }}=1[\because \cos ({90}^{\circ }-\theta )=\sin \theta ]$
$\left(ii\right)\cos {48}^{\circ }-\sin {42}^{\circ }=\cos ({90}^{\circ }-{42}^{\circ })-\sin {42}^{\circ }=\sin {42}^{\circ }-\sin {42}^{\circ }=0$