The value of the angle ${\tan }^{-1}\left(\tan {65}^{\circ }-2\tan {40}^{\circ }\right)$ in degrees is equal to

$\begin{array}{r}\tan \left({45}^{\circ }+{20}^{\circ }\right)-2\tan {40}^{\circ }\\ \frac{1+\tan {20}^{\circ }}{1-\tan {20}^{\circ }}-\frac{4\tan {20}^{\circ }}{1-{\tan }^2{20}^{\circ }}\end{array}$

$\begin{array}{l}=\frac{\left(1-\tan {20}^{\circ }\right)}{\left(1+\tan {20}^{\circ }\right)}\\ =\tan \left({45}^{\circ }-{20}^{\circ }\right)=\tan {25}^{\circ }\\ \therefore {\tan }^{-1}\left(\tan {25}^{\circ }\right)={25}^{\circ }\end{array}$