Question

sin (45° + θ) − cos (45° − θ) = ?
(a) 2 sin θ
(b) 2 cos θ
(c) 0
(d) 1

Answer 1

$$\begin{gathered} (c)\; 0 \\ \text{We have:} \\ [\sin ({45}^0+\theta )-\cos ({45}^0-\theta )] \\ =[\sin \{{90}^0-({45}^0-\theta )\}-\cos ({45}^0-\theta )] \\ \text{=[cos}({45}^0-\theta )-\cos ({45}^0-\theta )\text{]}[\because \sin ({90}^0-\theta )=co\mathrm{s}\theta ] \\ =0 \end{gathered}$$