Question

$\frac{\tan 35°}{\cot 55°}+\frac{\cot 78°}{\tan 12°}=?$
(a) 0
(b) 1
(c) 2
(d) None of these

Answer 1

$$\begin{gathered} \text{(c) 2} \\ \text{We have:} \\ \frac{\tan {35}^0}{\cot {55}^0}+\frac{\cot {78}^0}{\tan {12}^o} \\ =\frac{\tan {35}^0}{\cot ({90}^0-{35}^0)}+\frac{\cot ({90}^0-{12}^0)}{\tan {12}^0} \\ =\frac{\tan {35}^0}{\tan {35}^0}+\frac{\tan {12}^0}{\tan {12}^0}[\because \cot ({90}^0-\theta )=\tan \theta ] \\ =1+1=2 \end{gathered}$$