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  4. If sin (α+β)=1, sin (α-β)=(1/2), then tan (α+2 β) tan (2 α+β), α, β ∈(0, π / 2), is equal to

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Q. If $\sin (\alpha+\beta)=1, \sin (\alpha-\beta)=\frac{1}{2}$, then $\tan (\alpha+2 \beta)$ $\tan (2 \alpha+\beta), \alpha, \beta \in(0, \pi / 2)$, is equal to

Trigonometric Functions

Solution:

$\sin (\alpha+\beta)=1$
or $\alpha+\beta=\frac{\pi}{2} \sin (\alpha-\beta)=\frac{1}{2}$
or $(\alpha-\beta)=\frac{\pi}{6}$
Solving, we get $\alpha=\pi / 3$ and $\beta=\pi / 6$.
Now $\tan (\alpha+2 \beta) \tan (2 \alpha+\beta)$
$=\tan \left(\frac{\pi}{3}+\frac{\pi}{3}\right) \tan \left(\frac{2 \pi}{3}+\frac{\pi}{6}\right)$
$=\tan \frac{2 \pi}{3} \tan \frac{5 \pi}{6}$
$=\left(-\cot \frac{\pi}{3}\right)\left(-\cot \frac{\pi}{6}\right)$
$=\left(-\frac{1}{\sqrt{3}}\right)(-\sqrt{3})$
$=1$