For the equation 

$1-2x-x^2={\tan }^2\left(x+y\right)+{\cot }^2\left(x+y\right)$

$\begin{array}{l}1-2x-x^2={\tan }^2\left(x+y\right)+{\cot }^2\left(x+y\right)\\ \Rightarrow 1-2x-x^2-2={\tan }^2\left(x+y\right)+{\cot }^2\left(x+y\right)-2\\ \Rightarrow -{\left(x+1\right)}^2={\left[\tan \left(x+y\right)-\cot \left(x+y\right)\right]}^2\\ NowL.{\rm H}.S\le 0andR.{\rm H}.\delta \ge 0\\ \Rightarrow -{\left(x+1\right)}^2={\left[\tan \left(x+y\right)-\cot \left(x+y\right)\right]}^2=0\\ \text{⇒\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}x=-1and\tan \left(x+y\right)=\cot \left(x+y\right)\\ \text{\hspace{0.17em}⇒\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}x=-1and{\tan }^2\left(-1+y\right)=1\\ \Rightarrow x=-1and-1+y=n\pi \pm \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.:n\in z\\ \Rightarrow y=1+n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{4}\end{array}$