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已知$\frac{\sin^2\theta\cos\varphi-\cos^2\theta\sin\varphi}{\cos\theta\tan\alpha}=\frac{\sin^2\varphi \cos\theta-\cos^2\varphi\sin\theta}{\cos\varphi\tan\beta}=\cos(\theta+\varphi)$，求证$\frac{\sin^2\alpha \cos\beta -\cos^2\alpha \sin\beta }{\cos\alpha\tan\theta}=\frac{\sin^2\beta\cos\alpha-\cos^2\beta\sin\alpha}{\cos\beta\tan\varphi}=\cos(\alpha +\beta )$ |
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