Question

$\frac{\cos 12°}{\sin 78°}+\frac{\tan 23°}{\cot 67°}=?$
(a) 0
(b) 1
(c) 2
(d) $\frac{3}{2}$

Answer 1

$$\begin{gathered} \frac{\cos 12°}{\sin 78°}+\frac{\tan 23°}{\cot 67°}=\frac{\cos \left(90°-78°\right)}{\sin 78°}+\frac{\tan 23°}{\cot 67°} \\ =\frac{\sin 78°}{\sin 78°}+\frac{\tan 23°}{\cot 67°}\left(\because \cos \left(90°-\theta \right)=\sin \theta \right) \\ =1+\frac{\tan \left(90°-67°\right)}{\cot 67°} \\ =1+\frac{\cot 67°}{\cot 67°}\left(\because \tan \left(90°-\theta \right)=\cot \theta \right) \\ =1+1 \\ =2 \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{c}\right). \end{gathered}$$