Question

If cos ($\alpha +\beta$) = 0 , then sin $\left(\alpha -\beta \right)$ can be reduced to

Answer 1

It is given that,
$$\begin{gathered} \cos \left(\alpha +\beta \right)=0 \\ \Rightarrow \cos \left(\alpha +\beta \right)=\cos 90°\left(\cos 90°=0\right) \\ \Rightarrow \alpha +\beta =90° \\ \Rightarrow \alpha =90°-\beta \\ \mathrm{Now},\mathrm{put}\alpha =90°-\beta \mathrm{in}\sin \left(\alpha -\beta \right). \\ \therefore \sin \left(\alpha -\beta \right) \\ =\sin \left(90°-\beta -\beta \right) \\ =\sin \left(90°-2\beta \right) \\ =\cos 2\beta \left[\sin \left(90°-\theta \right)=\cos \theta \right] \end{gathered}$$
Hence, the correct answer is option (b).