If \(\sin(\alpha+\beta) = 1\) and \(\sin(\alpha-\beta) = \frac{1}{2}\), then \(\tan(\alpha+2\beta) \cdot \tan(2\alpha+\beta)\) equals:
A. 1
B. -1
C. 0
D. None of the above
Answer (2)
\alpha;\ \beta\in(0^o;\ 90^o)\\\\sin(\alpha+\beta)=1\to sin(\alpha+\beta)=sin90^o\to\alpha+\beta=90^o\\\\sin(\alpha-\beta)=\frac{1}{2}\to sin(\alpha-\beta)=sin30^o\to\alpha-\beta=30^o\\\\ +\left\{\begin{array}{ccc}\alpha+\beta=90^o\\\alpha-\beta=30^o\end{array}\right\\---------\\.\ \ \ \ \ \ \ 2\alpha=120^o\ \ \ /:2\\.\ \ \ \ \ \ \ \ \ \alpha=60^o\\\\60^o+\beta=90^o\ \ \ /-60^o\\\beta=90^o-60^o\\\beta=30^o
t an ( α + 2 β ) ⋅ t an ( 2 α + β ) = t an ( 6 0 o + 2 ⋅ 3 0 o ) ⋅ t an ( 2 ⋅ 6 0 o + 3 0 o ) = t an 12 0 o ⋅ t an 15 0 o = t an ( 18 0 o − 6 0 o ) ⋅ t an ( 18 0 o − 3 0 o ) = − t an 6 0 0 ⋅ ( − t an 3 0 o ) = − 3 ⋅ ( − 3 3 ) = 3 3 = 1 A n s w er : A
The calculated value of tan ( α + 2 β ) ⋅ tan ( 2 α + β ) is 1 , based on the sine values given. Therefore, the correct answer is option A.
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