Tangent Sum Formula Application Given that and , find .

Trigonometry Solution – Problem 1: find

Tangent Sum Formula Application

Given that $\tan\alpha = 2$ and $2\sin\alpha = \cos(\alpha-\beta)\sin\beta$ , find $\tan\beta$ .

Since $\sin\alpha = \sin[(\alpha-\beta)+\beta]$ :

$\sin\alpha = \sin(\alpha-\beta)\cos\beta + \cos(\alpha-\beta)\sin\beta$

From $2\sin\alpha = \cos(\alpha-\beta)\sin\beta$ :

$2\sin(\alpha-\beta)\cos\beta + 2\cos(\alpha-\beta)\sin\beta = \cos(\alpha-\beta)\sin\beta$

Therefore:

$2\sin(\alpha-\beta)\cos\beta = -\cos(\alpha-\beta)\sin\beta$

$2\tan(\alpha-\beta) = -\tan\beta$

Since $\tan\alpha = 2$ , we have $\tan(\alpha-\beta) = \frac{3}{2}\tan\beta$ .

Using the tangent difference formula:

$\tan(\alpha-\beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha\cdot\tan\beta} = \frac{2 - \tan\beta}{1 + 2\tan\beta}$

Solving: $2 - \tan\beta = 3 + 6\tan\beta$ , we get $\tan\beta = \frac{1}{7}$ .

$\frac{1}{7}$

Correct identity setup (2 pts): choose an appropriate sum/difference, double-angle, or auxiliary-angle idea and set up the key equation(s).
Correct algebra / trig simplification (2 pts): transform expressions without sign mistakes.
Solve for target quantity (2 pts): isolate the requested value and handle any constraints if needed.
Final answer (1 pt): clearly state the result in the required form.