If  $\alpha ,\beta ,\gamma$ are acute angles and  $\cos \theta =\frac{\sin \beta }{\sin \alpha },\cos \varphi =\frac{\sin \gamma }{\sin \alpha }and\cos (\theta -\varphi )=\sin \beta \sin \gamma ,$ then the value of ${\tan }^2\alpha -{\tan }^2\beta -{\tan }^2\gamma$ is equal to

$$\begin{gathered} \cos \theta \cos \varphi +\sin \theta \sin \varphi =\sin \beta \sin \gamma \Rightarrow {\sin }^2\theta {\sin }^2\varphi ={(\cos \theta \cos \varphi -\sin \beta \sin \gamma )}^2 \\ \Rightarrow (1-\frac{{\sin }^2\beta }{{\sin }^2\alpha })(1-\frac{{\sin }^2\gamma }{{\sin }^2\alpha })={(\frac{\sin \beta \sin \gamma }{{\sin }^2\alpha }-\sin \beta \sin \gamma )}^2 \\ \Rightarrow ({\sin }^2\alpha -{\sin }^2\beta )({\sin }^2\alpha -{\sin }^2\gamma )={\sin }^2\beta {\sin }^2\gamma {(1-{\sin }^2\alpha )}^2 \\ \Rightarrow {\sin }^4\alpha (1-{\sin }^2\beta {\sin }^2\gamma )-{\sin }^2\alpha ({\sin }^2\beta +{\sin }^2\gamma -2{\sin }^2\beta {\sin }^2\gamma )=0 \\ \therefore {\sin }^2\alpha =\frac{{\sin }^2\beta +{\sin }^2\gamma -2{\sin }^2\beta .{\sin }^2\gamma }{1-{\sin }^2\beta .{\sin }^2\gamma } and {\cos }^2\alpha =\frac{1-{\sin }^2\beta -{\sin }^2\gamma +{\sin }^2\beta {\sin }^2\gamma }{1-{\sin }^2\beta .{\sin }^2\gamma } \\ \Rightarrow {\tan }^2\alpha =\frac{{\sin }^2\beta -{\sin }^2\beta {\sin }^2\gamma +{\sin }^2\gamma -{\sin }^2\beta {\sin }^2\gamma }{{\cos }^2\beta -{\sin }^2\gamma (1-{\sin }^2\beta )} \\ =\frac{{\sin }^2\beta {\cos }^2\gamma +{\cos }^2\beta {\sin }^2\gamma }{{\cos }^2\beta -{\sin }^2\gamma }={\tan }^2\beta +{\tan }^2\gamma \\ ={\tan }^2\alpha -{\tan }^2\beta -{\tan }^2\gamma =0 \end{gathered}$$