Question

Please solve this trigonometric equation!!​

Answer 1

Solution....(swipe left)

$L.H.S. = \frac{ \cos(2a )\cos(3a) - \cos(2a) \cos(7a) + \cos(a) \cos(10a) }{ \sin(4a) \sin( 3a) - \sin(2a) \sin(5a) + sin(4a) \sin(7a) } \\ = \frac{ \cos(5a) + \cos( a) - \cos(9a) - \cos(5a) + \cos(11a) + \cos(9a) }{ \cos(a) - \cos(7a) - \cos(3a) + \cos(7a) + \cos(3a) - \cos(11a) } \\ = \frac{ \cos(a) + \cos(11a) }{ \cos(a) - \cos(11a) } \\ = \frac{ \cos(6a - 5a) + \cos(6a + 5a) }{ \cos(6a - 5a) - \cos(6a + 5a) } \\ = \frac{2 \cos(6a) \cos(5a) }{2 \sin(6a) \sin(5a) } \\ = \cot(6a) \cot(5a) =R.H.S.(proved) \\ \\ formula \: used \: ..........\\ 2 \cos(a) \cos(b) = \cos(a - b) + \cos(a + b) \\ 2 \sin(a) \sin(b) = \cos(a - b) - \cos(a + b)$

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