मराठी

If Cos ( α + β = 0 , Then Sin ( α − β ) Can Be Reduced to

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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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पाठ 11: Trigonometric Identities - Exercise 11.4 [पृष्ठ ५९]

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आर.डी. शर्मा Mathematics [English] Class 10
पाठ 11 Trigonometric Identities
Exercise 11.4 | Q 33 | पृष्ठ ५९

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