已知$\alpha,\beta,\gamma$是三个互不相等的锐角,若$tan\alpha=\dfrac{\sin\beta\sin\gamma}{\cos\beta-\cos\gamma}$则$\tan\beta=$______

 

 

解答:

$\tan^2\alpha+1=\dfrac{\sin^2\beta\sin^2\gamma}{(\cos\beta-\cos\gamma)^2}+1$

$=\dfrac{(1-cos^2\beta)(1-cos^2\gamma)+(cos\beta-cos\gamma)^2}{(\cos\beta-\cos\gamma)^2}$
$=\dfrac{(1-cos\beta\cos\gamma)^2}{(\cos\beta-\cos\gamma)^2}$
故$cos\alpha=\dfrac{cos\beta-cos\gamma}{1-cos\beta\cos\gamma}$
解得$\cos\beta=\dfrac{cos\alpha+cos\gamma}{1+cos\alpha\cos\gamma}=\dfrac{cos\alpha-cos(\pi-\gamma)}{1-cos\alpha\cos(\pi-\gamma)}$
考虑到$\alpha$与$\beta$的对称性与$\gamma$与$\pi-\gamma$的对称性；
故$tan\beta=\dfrac{sin\alpha\sin(\pi-\gamma)}{cos\alpha-cos(\pi-\gamma)}=\dfrac{\sin\alpha\sin\gamma}{cos\alpha+\cos\gamma}$