Question

sec 70° sin 20° + cos 20° cosec 70° = ?
(a) 0
(b) 1
(c) −1
(d) 2

Answer 1

$$\begin{gathered} (d)\; 2 \\ \text{We have:} \\ {\text{sec70}}^0\sin {20}^0+\cos {20}^0\cos {\text{ec70}}^0 \\ =\frac{{\text{sin20}}^0}{{\text{cos70}}^0}+\frac{\cos {20}^0}{{\text{sin70}}^0} \\ =\frac{\sin {20}^0}{\cos ({90}^0-{20}^0)}+\frac{{\text{cos20}}^0}{\sin ({90}^0-{20}^0)} \\ =\frac{\sin {20}^0}{\sin {20}^0}+\frac{{\text{cos20}}^0}{{\text{cos20}}^0}\left[\begin{array}{l}\because \cos ({90}^0-\theta )=\sin \theta \text{and}\\ \sin ({90}^0-\theta )=\cos \theta \end{array}\right] \\ =1+1 \\ =2 \end{gathered}$$