Question

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

• A. $\cos \left(\beta \right)$
• B. $\cos \left(2\beta \right)$
• C. $\sin \left(\alpha \right)$
• D. $\sin \left(2\alpha \right)$

Answer 1

The correct option is B

$\cos \left(2\beta \right)$

Explanation for correct answer

According to the given details

$$\begin{gathered} \cos \left(\alpha +\beta \right)=0 \\ \therefore \alpha +\beta =90°[\because \cos \left(90°\right)=0] \\ \Rightarrow \alpha =90°-\beta \end{gathered}$$

Now,

$$\begin{gathered} \sin \left(\alpha -\beta \right) \\ =\sin \left(90°-\beta -\beta \right)[\because \alpha =90°-\beta ] \\ =\sin \left(90°-2\beta \right) \\ =\cos \left(2\beta \right)[\because \sin \left(90°-x\right)=\cos x] \end{gathered}$$

Hence, as $\sin \left(\alpha -\beta \right)=\cos \left(2\beta \right)$, so, the correct option is B.