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Prove That: Sin 3 a + Sin 5 a + Sin 7 a + Sin 9 a Cos 3 a + Cos 5 a + Cos 7 a + Cos 9 a = Tan 6 a - Mathematics

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प्रश्न

Prove that:

\[\frac{\sin 3A + \sin 5A + \sin 7A + \sin 9A}{\cos 3A + \cos 5A + \cos 7A + \cos 9A} = \tan 6A\]
योग
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उत्तर

Consider LHS: 
\[ \frac{\sin 3A + \sin 5A + \sin 7A + \sin 9A}{\cos 3A + \cos 5A + \cos 7A + \cos 9A}\]
\[ = \frac{\sin 3A + \sin 9A + \sin 5A + \sin 7A}{\cos 3A + \cos 9A + \cos 5A + \sin 7A}\]
\[ = \frac{2\sin \left( \frac{3A + 9A}{2} \right) \cos \left( \frac{3A - 9A}{2} \right) + 2\sin \left( \frac{5A + 7A}{2} \right) \cos \left( \frac{5A - 7A}{2} \right)}{2\cos \left( \frac{3A + 9A}{2} \right) \cos \left( \frac{3A - 9A}{2} \right) + 2\cos \left( \frac{5A + 7A}{2} \right) \cos \left( \frac{5A - 7A}{2} \right)}\]
\[ = \frac{2\sin 6A \cos \left( - 3A \right) + 2\sin 6A \cos \left( - A \right)}{2\cos 6A \cos \left( - 3A \right) + 2\cos 6A \cos \left( - A \right)}\]
\[ = \frac{2\sin 6A \cos 3A + 2\sin 6A \cos A}{2\cos 6A \cos 3A + 2\cos 6A \cos A}\]
\[ = \frac{2\sin 6A\left[ \cos 3A + \cos A \right]}{2\cos 6A\left[ \cos 3A + \cos A \right]}\]
\[ = \tan 6A\]
 = 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]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 8 Transformation formulae
Exercise 8.2 | Q 8.04 | पृष्ठ १८

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