Question:
$\frac{\cos 12^{\circ}}{\sin 78^{\circ}}+\frac{\tan 23^{\circ}}{\cot 67^{\circ}}=?$
(a) 0
(b) 1
(c) 2
(d) $\frac{3}{2}$
Solution:
$\frac{\cos 12^{\circ}}{\sin 78^{\circ}}+\frac{\tan 23^{\circ}}{\cot 67^{\circ}}=\frac{\cos \left(90^{\circ}-78^{\circ}\right)}{\sin 78^{\circ}}+\frac{\tan 23^{\circ}}{\cot 67^{\circ}}$
$=\frac{\sin 78^{\circ}}{\sin 78^{\circ}}+\frac{\tan 23^{\circ}}{\cot 67^{\circ}} \quad\left(\because \cos \left(90^{\circ}-\theta\right)=\sin \theta\right)$
$=1+\frac{\tan \left(90^{\circ}-67^{\circ}\right)}{\cot 67^{\circ}}$
$=1+\frac{\cot 67^{\circ}}{\cot 67^{\circ}} \quad\left(\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right)$
$=1+1$
$=2$
Hence, the correct option is (c).