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Prove That: Sin a + Sin 3 a + Sin 5 a Cos a + Cos 3 a + Cos 5 a = Tan 3 a

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प्रश्न

Prove that:

\[\frac{\sin A + \sin 3A + \sin 5A}{\cos A + \cos 3A + \cos 5A} = \tan 3A\]

 

योग
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उत्तर

Consider LHS: 
\[ \frac{\sin A + \sin 3A + \sin 5A}{\cos A + \cos 3A + \cos 5A}\]
\[ = \frac{\sin A + \sin 5A + \sin 3A}{\cos A + \cos 5A + \cos 3A}\]
\[ = \frac{2\sin \left( \frac{A + 5A}{2} \right) \cos \left( \frac{A - 5A}{2} \right) + \sin 3A}{2\cos \left( \frac{A + 5A}{2} \right) \cos \left( \frac{A - 5A}{2} \right) + \cos 3A}\]
\[ \]
\[ = \frac{2 \sin 3A \cos \left( - 2A \right) + \sin 3A}{2\cos 3A \cos \left( - 2A \right) + \cos 3A}\]
\[ = \frac{2\sin 3A \cos 2A + \sin 3A}{2\cos 3A \cos 2A + \cos 3A}\]
\[ = \frac{\sin 3A \left[ 2\cos 2A + 1 \right]}{\cos 3A \left[ 2\cos 2A + 1 \right]}\]
\[ = \tan 3A\]
= RHS
Hence, RHS = LHS .

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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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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 8 Transformation formulae
Exercise 8.2 | Q 8.01 | पृष्ठ १८

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