$\color{Red}{\texttt{Find all the triangle ABC}}$ whose angles satisfies $$2\left (1+tan^2\frac{C}{2} \right )\left [ cos^2\left (\frac{13\pi }{2}+\frac{B}{2} \right )+\sqrt{2}sin\frac{B}{2}sin\frac{C}{2} \right ]+tan^2\frac{C}{2}=1\quad (1)$$ This is my try: $(1)\displaystyle \iff 2\left [ \cos^2\left (\frac{\pi }{2}+\frac{B}{2} \right )+\sqrt{2}\sin\frac{B}{2}\sin\frac{C}{2} \right ]=\frac{1-\tan^2\frac{C}{2}}{1+\tan^2\frac{C}{2}}\\\iff 2\left [ \sin^2\frac{B}{2}+\sqrt2\sin \frac B2\sin\frac C2 \right ]=\cos C\\\iff 2\left [ \frac{1-\cos B}{2}+\sqrt2\sin \frac B2\sin\frac C2 \right ]=\cos C\\\iff 1+2\sqrt{2}\sin\frac{B}{2}\sin\frac{C}{2}=\cos B+\cos C\\\iff 1+\sqrt{2}\left ( \cos\frac{B-C}{2}-\sin\frac{A}{2} \right )=2\sin \frac A2\cos\frac{B-C}{2}$
Then I do not know how to continue
$$1+2\sqrt2 \sin {\frac B 2} \sin {\frac C 2}=\cos B+\cos C \Leftrightarrow$$ $$8( \sin {\frac B 2} \sin {\frac C 2})^2=(\cos B+\cos C-1)^2 \Leftrightarrow$$ $$2(1-\cos B)(1-\cos C)=(\cos B+\cos C-1)^2 \Leftrightarrow$$ $$\cos ^2 B+\cos ^2C=1 \Leftrightarrow$$ $$\cos B=\sin C$$ so the triangle is right triangle.