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

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प्रश्न

Prove that:

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

 

संक्षेप में उत्तर
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उत्तर

\[\text{ RHS }= \frac{\tan(x - b) - \tan(x - a)}{\sin(a - b)}\]
\[ = \frac{\frac{\sin(x - b)}{\cos(x - b)} - \frac{\sin(x - a)}{\cos(x - a)}}{\sin(a - b)}\]
\[ = \frac{\sin(x - b) \cos(x - a) - \sin(x - a) \cos(x - b)}{\sin(a - b) \cos(x - a) \cos(x - b)}\]
\[ = \frac{\sin(x - b - x + a)}{\sin(a - b) \cos(x - a) \cos(x - b)} ( \text{ Using }\sin(A - B) = \sin A\cos B - \cos A\sin B)\]
\[ = \frac{\sin(a - b)}{\sin(a - b) \cos(x - a) \cos(x - b)}\]
\[ = \frac{1}{\cos(x - a) \cos(x - b)} \]
 = LHS
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 29.3 | पृष्ठ २१

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