Type of Triangle 1

$\frac{D}{4}=\cos^2 B+(\sin C+\cos A)(\sin C-\cos A)=0$ $\cos^2 B+\sin^2 C-\cos^2 A=0$ Due to the Pythagorean Trigonometric Identity $\sin^2 \theta + \cos^2 \theta = 1$ $1-\sin^2 B+\sin^2 C-(1-\sin^2 A)=0$ $\sin^2 A-\sin^2 B+\sin^2 C=0$ Due to the Law of Sines $\sin A=\frac{a}{2R}$ $\sin B=\frac{b}{2R}$ $\sin C=\frac{c}{2R}$ Where $R$ represents the radius of the circumscribed circle of $\bigtriangleup ABC$ $\left(\frac{a}{2R}\right)^2-\left(\frac{b}{2R}\right)^2+\left(\frac{c}{2R}\right)^2=0$ $\frac{a^2-b^2+c^2}{4R^2}=0$ $\therefore a^2+c^2=b^2 \hspace{1cm} (\because R \neq 0)$ Therefore the triangle $\bigtriangleup ABC$ is a right triangle where $\angle B$ is right angle.