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प्रश्न
Prove that:
`(cos 4x + cos 3x + cos 2x)/(sin 4x + sin 3x + sin 2x)` = cot 3x
उत्तर
LHS = `(cos 4x + cos 3x + cos 2x)/(sin 4x + sin 3x + sin 2x)`
`= (2 cos ((4x + 2x)/2) * cos ((4x - 2x)/2) + cos 3x)/(2sin ((4x + 2x)/2) * cos((4x - 2x)/2) + sin 3x)`
`[because cos "C" + cos "D" = 2cos (("C + D")/2) cos (("C - D")/2) and sin "C" + sin "D" = 2 sin (("C + D")/2) cos (("C - D")/2)]`
`= (cos 3x * cos x + cos 3x)/(sin 3x * cos x + sin 3x)`
`= (cos 3x cancel((cosx + 1)))/(sin 3x cancel((cos x + 1)))`
= cot 3x = RHS
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