《2025年新课程学习指导高中数学必修第二册北师大版》

[典例3]　(1)化简:$\frac{1}{\tan\theta + 1} + \frac{1}{\tan\theta - 1} =$________.
　　(2)求证:$\frac{\sqrt{3}\tan12^{\circ} - 3}{\sin12^{\circ}(4\cos^{2}12^{\circ} - 2)} = - 4\sqrt{3}$

答案：
(1)−tan20°
(2)证明:左边=$\frac{\sqrt{3}\sin12^{\circ}-3\cos12^{\circ}}{2\sin12^{\circ}(2\cos^{2}12^{\circ}-1)}$
　=$\frac{2\sqrt{3}(\frac{1}{2}\sin12^{\circ}-\frac{\sqrt{3}}{2}\cos12^{\circ})}{2\sin12^{\circ}\cos12^{\circ}\cos24^{\circ}}$
　=$\frac{2\sqrt{3}\sin(12^{\circ}-60^{\circ})}{\sin24^{\circ}\cos24^{\circ}}$
　=$\frac{-2\sqrt{3}\sin48^{\circ}}{\sin48^{\circ}}=-4\sqrt{3}$=右边,所以原等式成立.

　求证:(1)$\cos^{2}(A + B) - \sin^{2}(A - B) = \cos2A\cos2B$;
　(2)$\frac{1 + \sin\alpha - \cos\alpha}{1 + \sin\alpha + \cos\alpha} + \frac{1 + \cos\alpha + \sin\alpha}{1 - \cos\alpha + \sin\alpha} = \frac{2}{\sin\alpha}$.

答案： 证明:
(1)左边=$\frac{1+\cos(2A + 2B)}{2}-\frac{1-\cos(2A - 2B)}{2}$
　=$\frac{\cos(2A + 2B)+\cos(2A - 2B)}{2}$
　=$\frac{1}{2}$(cos2Acos2B−sin2Asin2B+cos2Acos2B+sin2Asin2B)
　=cos2Acos2B=右边，所以原等式成立.
(2)左边=$\frac{\sin\frac{\alpha}{2}(\cos\frac{\alpha}{2}+\sin\frac{\alpha}{2})}{2\cos\frac{\alpha}{2}(\cos\frac{\alpha}{2}+\sin\frac{\alpha}{2})}+\frac{\cos\frac{\alpha}{2}(\cos\frac{\alpha}{2}+\sin\frac{\alpha}{2})}{2\sin\frac{\alpha}{2}(\sin\frac{\alpha}{2}+\cos\frac{\alpha}{2})}$
　=$\frac{\sin\frac{\alpha}{2}}{2\cos\frac{\alpha}{2}}+\frac{\cos\frac{\alpha}{2}}{2\sin\frac{\alpha}{2}}$
　=$\frac{\sin^{2}\frac{\alpha}{2}+\cos^{2}\frac{\alpha}{2}}{2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}=\frac{1}{\sin\alpha}=\frac{2}{2\sin\alpha}$=右边，所以原等式成立.