Question

sin θ cos (90° − θ) + cos θ sin (90° − θ) = ?
(a) 0
(b) 1
(c) 2
(d) $\frac{3}{2}$

Answer 1

(b) 1

$$\begin{gathered} \text{We have:} \\ \sin \theta \cos ({90}^0-\theta )+\cos \theta \sin ({90}^0-\theta ) \\ =\sin \theta .\sin \theta +\cos \theta .\cos \theta [\because \cos ({90}^0-\theta )=\sin \theta \text{and sin}({90}^0-\theta )=\cos \theta ] \\ ={\sin }^2\theta +{\cos }^2\theta \\ =1 \end{gathered}$$