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If 2 sin( + { }{ 3 } ) =( - { }{ 6 } ) , prove that

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If 2 sin $(θ + \frac{π}{3}) = cos (θ - \frac{π}{6}),$ prove that $tan θ + 3 = 0$

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If 2 sin

(θ + \frac{π}{3}) = cos (θ - \frac{π}{6}),

prove that

tan θ + 3 = 0

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12 mins ago

Text solutionVerified

= 2 [sin θ cos \frac{π}{3} + cos θ sin \frac{π}{3}] = (cos θ \cdot cos \frac{π}{6} + sin θ sin \frac{π}{6}) = 2 [sin θ (\frac{1}{2}) + cos θ (\frac{3}{2})] = [\frac{3}{2} cos θ + \frac{1}{2} sin θ] = sin θ + 3 cos θ = \frac{3}{2} cos θ + \frac{1}{2} sin θ = 3 cos θ - \frac{3}{2} cos θ = \frac{1}{2} sin θ - sin θ = 3 (\frac{2 cos - cos θ}{2}) = \frac{sin θ - 2 sin θ}{2} = 3 cos θ = - sin θ = \frac{- sin θ}{cos θ} = 3 = - tan θ = 3 \Rightarrow 3 + tan θ = 0

Hence, proved.

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Question Text	If 2 sin $(θ + \frac{π}{3}) = cos (θ - \frac{π}{6}),$ prove that $tan θ + 3 = 0$
Answer Type	Text solution:1
Upvotes	150