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Answer 1

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Step-by-step explanation:

$to \: prove \: = \\ \frac{sin \: 40 - sin \: 20}{cos \: 220 - cos \: 200} = \sqrt{3} \\ \\ so \: let \\ LHS = \frac{sin \: 40 - sin \: 20}{cos \: 220 - cos \: 200} \\ \\ RHS = \sqrt{3}$

$considering \: LHS \\ = \frac{sin \: 40 - sin \: 20}{cos \: 220 - cos \: 200} \\ \\ = \frac{sin \: (30 + 10) - sin \: (30 - 10)}{cos \: (210 + 10) - cos \: (210 - 10)} \\ \\ so \: here \: \\ sin \: (30 + 10) \: \: is \: in \: form \: sin \: (A + B ) \\ sin \: (30 - 10) \: is \: in \: form \: sin \: (A - B) \\ cos \: (210 + 10) \: is \: in \: form \: cos \: (A + B) \\ cos \: (210 - 10) \: is \: in \: form \: cos \: (A - B)$

$so \: we \: know \: that \\ sin \: (A + B) = sin \: A.cos \: B \: + cos \: A.sin \: B \\ \\ sin \: (A - B) = sin \: A.cos \: B \: - \: cos \: A.sin \: B \\ \\ cos \: (A + B) = cos \: A.cos \: B \: - \: sin \: A.sin \: B \\ \\ cos \: (A - B) = cos \: A.cos \: B + sin \: A. sin \: B$

$hence \: accordingly \: \\ = \frac{sin \: 30.cos \: 10 \: + \: cos \:30.sin \: 10 \: - \: (sin \: 30.cos \: 10 \: - \: cos \: 30.sin \: 10) }{cos \: 210.cos \: 10 - sin \: 210.sin \: 10 \: - \: (cos \: 210.cos \: 10 \: + sin \: 210.sin \: 10)} \\ \\ = \frac{sin \: 30.cos \: 10 + cos \: 30.sin \: 10- sin \: 30.cos \: 10 + cos \: 30.sin \: 10}{cos \: 210.cos \: 10 - sin \: 210.sin \: 10 - cos \: 210.cos \: 10 - sin \: 210.sin \: 10 } \\ \\ = \frac{cos \: 30.sin \: 10 + cos \: 30.sin \: 10}{ - sin \: 210.sin \: 10 - sin \: 210.sin \: 10} \\ \\ = \frac{2.cos \: 30.sin \: 10}{ - 2.sin \: 210.sin \: 10} \\ \\ = \frac{cos \: 30}{ - sin \: 210} \\ \\ so \: sin \: 210 \: can \: be \: written \: as \: sin \: (270 - 60) \: \\ ie \: sin \: ( \frac{3\pi}{2} - \frac{\pi}{3} ) \\ \\ so \: we \: know \: that \\ whenever \: there \: is \: \frac{\pi}{2} \: in \: any \: trigonometric \: ratio \\ then \: that \: ratio \: changes \: to \: its \: complementary \: ratio \\ \\ hence \: sin \: 210 = cos \: 60$

$so \: thus \: then \\ we \: know \: that \\ cos \: 30 = \frac{ \sqrt{3} }{2} \\ \\ = \frac{ \frac{ \frac{ \sqrt{3} }{2} }{ - ( - 1)} }{2} \\ \\ = \frac{ \frac{ \frac{ \sqrt{3} }{2} }{1} }{2} \\ \\ = \sqrt{3}$

$hence \: LHS=RHS \\ thus \: proved$

$thus \: then \\ we \: know \: that \: \\ cos \: 60 = \frac{1}{2} \\ \\ but \: as \: 210 \: lies \: in \: quadrant \: 3 \\ we \: know \: that \: except \: tan \: and \: cot \: function \\ all \: others \: are \: negative \\ \\ hence \: cos \: 210 = - \frac{1}{2}$