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प्रश्न
If cos (\[\alpha + \beta\]= 0 , then sin \[\left( \alpha - \beta \right)\] can be reduced to
बेरीज
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उत्तर
It is given that,
\[\cos\left( \alpha + \beta \right) = 0\]
\[ \Rightarrow \cos\left( \alpha + \beta \right) = \cos90° \left( \cos90° = 0 \right)\]
\[ \Rightarrow \alpha + \beta = 90° \]
\[ \Rightarrow \alpha = 90°- \beta\]
\[\text{ Now, put }\alpha = 90°- \beta in \sin\left( \alpha - \beta \right) . \]
\[ \therefore \sin\left( \alpha - \beta \right)\]
\[ = \sin\left( 90° - \beta - \beta \right)\]
\[ = \sin\left( 90°- 2\beta \right) \]
\[ = \cos2\beta \left[ \sin\left( 90° - \theta \right) = \cos\theta \right]\]
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