If $ \cos (\alpha+\beta)=0 $, then $ \sin (\alpha-\beta) $ can be reduced to
(A) $ \cos \beta $
(B) $ \cos 2 \beta $
(C) $ \sin \alpha $
(D) $ \sin 2 \alpha $

AcademicMathematicsNCERTClass 10

Given:

\( \cos (\alpha+\beta)=0 \)

To do:

We have to find the value of \( \sin (\alpha-\beta) \)

Solution:  

$\cos (\alpha+\beta) =0$

$=\cos 90^{\circ}$          [Since $\cos 90^{\circ}=0$]

This implies,

$\alpha+\beta =90^{\circ}$

$\alpha =90^{\circ}-\beta$

$\sin\ (\alpha-\beta) =\sin (90^{\circ}-\beta-\beta)$

$=\sin\ (90^{\circ}-2 \beta)$

$=\cos 2 \beta$           [Since $\sin\ (90^{\circ}-\theta)=\cos \theta$]

Hence, $\sin (\alpha-\beta)$ can be reduced to $\cos 2 \beta$.

raja
Updated on 10-Oct-2022 13:28:52

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