Question

$\left(\frac{\sin 49°}{\cos 41°}-\frac{\cos 17°}{\sin 73°}\right)=?$
(a) 1
(b) 0
(c) −1
(d) Cannot be calculated

Answer 1

$$\begin{gathered} (b)\; 0 \\ \text{We have:} \\ (\frac{\sin {49}^0}{\cos {41}^0}-\frac{\cos {17}^0}{\sin {73}^0}) \\ =[\frac{\sin {49}^0}{\cos ({90}^0-{49}^0)}-\frac{\cos ({90}^0-{73}^0)}{\sin {73}^0}] \\ =(\frac{\sin {49}^0}{\sin {49}^0}-\frac{\sin {73}^0}{\sin {73}^0})[\because \cos ({90}^0-\theta )=\sin \theta ] \\ =1-1 \\ =0 \end{gathered}$$