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Prove That: 1 Sin ( X − a ) Cos ( X − B ) = Cot ( X − a ) + Tan ( X − B ) Cos ( a − B )

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प्रश्न

Prove that:

\[\frac{1}{\sin \left( x - a \right) \cos \left( x - b \right)} = \frac{\cot \left( x - a \right) + \tan \left( x - b \right)}{\cos \left( a - b \right)}\]

 

संक्षेप में उत्तर
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उत्तर

\[\text{ RHS }= \frac{\cot(x - a) + \tan(x - b)}{\cos(a - b)} \]
\[ = \frac{\frac{\cos(x - a)}{\sin(x - a)} + \frac{\sin(x - b)}{\cos(x - b)}}{\cos(a - b)}\]
\[ = \frac{\cos(x - b) \cos(x - a) + \sin(x - a) \sin(x - b)}{\cos(a - b) \sin(x - a) \cos(x - b)}\]
\[ = \frac{\cos(x - b - x + a)}{\cos(a - b) \sin(x - a) \cos(x - b)} (\text{ Using }\cos(A - B) = \cos A \cos b B + \sin A \sin B)\]
\[ = \frac{\cos(a - b)}{\cos(a - b) \sin(x - a) \cos(x - b)}\]
\[ = \frac{1}{\sin(x - a) \cos(x - b)} \]
= RHS
Hence proved .

 

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अध्याय 7: Values of Trigonometric function at sum or difference of angles - Exercise 7.1 [पृष्ठ २१]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 7 Values of Trigonometric function at sum or difference of angles
Exercise 7.1 | Q 29.2 | पृष्ठ २१

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