If \sin\ (\alpha+\beta)=1, \sin\ (\alpha-\beta)=\dfrac {1}{2}, then \tan\ (\alpha+2\beta)\tan\ (2\alpha+\beta)=

1

- 1

zero

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If \sin\ (\alpha+\beta)=1, \sin\ (\alpha-\beta)=\dfrac {1}{2}, then \tan\ (\alpha+2\beta)\tan\ (2\alpha+\beta)=

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Text solutionVerified

Given

sin (α + β) = 1 ⟹ α + β = 9 0^{\circ} \dots (1)

sin (α - β) = \frac{1}{2} ⟹ α - β = 3 0^{\circ} \dots (2)

(1) + (2) ⟹ 2 α = 12 0^{\circ} ⟹ α = 6 0^{\circ}

⟹ β = 9 0^{\circ} - α = 9 0^{\circ} - 6 0^{\circ} = 3 0^{\circ}

tan (α + 2 β) tan (2 α + β) = tan (6 0^{\circ} + 6 0^{\circ}) tan (12 0^{\circ} + 3 0^{\circ}) = tan 12 0^{\circ} tan 15 0^{\circ} = (- 3) \times (- \frac{1}{3}) = 1

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Question Text	If \sin\ (\alpha+\beta)=1, \sin\ (\alpha-\beta)=\dfrac {1}{2}, then \tan\ (\alpha+2\beta)\tan\ (2\alpha+\beta)=
Answer Type	Text solution:1
Upvotes	150