The value of $\frac{e^{-{\displaystyle \frac{\pi }{4}}}+{\int }_0^{{\displaystyle \frac{\pi }{4}}}{e}^{-x}\tan { }^{50}xdx}{{\int }_0^{{\displaystyle \frac{\pi }{4}}}{e}^{-x}({\tan }^{49}x+{\tan }^{51}x)dx}$ 

Let,

$I=\frac{e^{-{\displaystyle \frac{\pi }{4}}}+{\int }_0^{{\displaystyle \frac{\pi }{4}}}{e}^{-x}\tan { }^{50}xdx}{{\int }_0^{{\displaystyle \frac{\pi }{4}}}{e}^{-x}\left({\tan }^{49}x+{\tan }^{51}x\right)dx}$

Now let $I_1={\int }_0^{\frac{\pi }{4}}{e}^{-x}({\tan }^{49}x+{\tan }^{51}x)dx$

Now solving, $I_2={\int }_0^{\frac{\pi }{4}}{e}^{-x}{\tan }^{50}xdx$ using byparts we get,

$I_2={\int }_0^{\frac{\pi }{4}}{e}^{-x}{\tan }^{50}xdx$

$\Rightarrow I_2={\left[-e^{-x}{\tan }^{50}x\right]}_0^{\frac{\pi }{4}}-{\int }_0^{\frac{\pi }{4}}50{\tan }^{49}x{\sec }^{2}x\left(-e^{-x}\right)dx$

$\Rightarrow I_2=-e^{-\frac{\pi }{4}}+{\int }_0^{\frac{\pi }{4}}50{\tan }^{49}x\left(1+{\tan }^{2}x\right)e^{-x}dx$

$\Rightarrow I_2=-e^{-\frac{\pi }{4}}+50{\int }_0^{\frac{\pi }{4}}\left({\tan }^{49}x+{\tan }^{51}x\right)e^{-x}dx$

$\Rightarrow I_2=-e^{-\frac{\pi }{4}}+50I_1$

$\Rightarrow {\int }_0^{\frac{\pi }{4}}{e}^{-x}{\tan }^{50}xdx+e^{-\frac{\pi }{4}}=50I_1$

Now putting the given integral $I$ we get,

$I=\frac{e^{-{\displaystyle \frac{\pi }{4}}}+{\int }_0^{{\displaystyle \frac{\pi }{4}}}{e}^{-x}\tan { }^{50}xdx}{{\int }_0^{{\displaystyle \frac{\pi }{4}}}{e}^{-x}\left({\tan }^{49}x+{\tan }^{51}x\right)dx}$

$\Rightarrow I=\frac{50I_1}{I_1}=50$