UPSC CDS 2 2023 Math Paper

Concept:Use algebraic identities to factor and simplify the given rational expression.Explanation:Start with the expression:$\frac{x^{2}-y^{2}-z^{2}-2yz}{x^{2}+y^{2}-z^{2}+2xy}+\frac{x^{2}-y^{2}-z^{2}-2yz}{x^{2}-y^{2}+z^{2}-2xz}$Rewrite each numerator by grouping the $y^{2}+z^{2}+2yz$ as $(y+z)^{2}$:Numerator: $x^{2}-(y^{2}+z^{2}+2yz)=x^{2}-(y+z)^{2}$First denominator: $x^{2}+y^{2}+2xy-z^{2}=(x+y)^{2}-z^{2}$Second denominator: $x^{2}+z^{2}-2xz-y^{2}=(x-z)^{2}-y^{2}$Now the expression becomes:$\frac{x^{2}-(y+z)^{2}}{(x+y)^{2}-z^{2}}+\frac{x^{2}-(y+z)^{2}}{(x-z)^{2}-y^{2}}$Apply the difference of squares identity $a^{2}-b^{2}=(a-b)(a+b)$ to each term:First fraction: $\frac{(x-(y+z))(x+(y+z))}{(x+y-z)(x+y+z)}=\frac{(x-y-z)(x+y+z)}{(x+y-z)(x+y+z)}$Second fraction: $\frac{(x-(y+z))(x+(y+z))}{(x-z-y)(x-z+y)}=\frac{(x-y-z)(x+y+z)}{(x-y-z)(x+y-z)}$Cancel common factors in each fraction:First fraction simplifies to $\frac{x-y-z}{x+y-z}$.Second fraction simplifies to $\frac{x+y+z}{x+y-z}$.Add the two simplified fractions since they share the same denominator $x+y-z$:$\frac{x-y-z+x+y+z}{x+y-z}=\frac{2x}{x+y-z}$Answer:$\frac{2x}{x+y-z}$