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Prove That: Sin a + Sin 3 a Cos a − Cos 3 a = Cot a

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प्रश्न

Prove that:

\[\frac{\sin A + \sin 3A}{\cos A - \cos 3A} = \cot A\]

 

योग
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उत्तर

Consider LHS: 
\[ \frac{\sin A + \sin 3A}{\cos A - \cos 3A}\]
\[ = \frac{2\sin \left( \frac{A + 3A}{2} \right) \cos \left( \frac{A - 3A}{2} \right)}{2\sin \left( \frac{A + 3A}{2} \right) \sin \left( \frac{3A - A}{2} \right)} \left\{ \because \sin A + \sin B = 2\sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right), and \cos A - \cos B = 2\sin \left( \frac{A + B}{2} \right) cos \left( \frac{B - A}{2} \right) \right\}\]
\[ = \frac{\sin 2A \cos \left( - A \right)}{\sin 2A \sin A}\]
\[ = \frac{\sin 2A \cos A}{\sin 2A \sin A}\]
\[ = \cot A\]
= RHS
Hence, LHS = RHS .

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Transformation Formulae
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Transformation formulae - Exercise 8.2 [पृष्ठ १८]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 8 Transformation formulae
Exercise 8.2 | Q 7.1 | पृष्ठ १८

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