UPSC NDA I 2024 Mathematics Paper

Given triangle $A B C$ with $\angle C=60^{\circ}$, we are required to find the value of $\frac{\cos A+\cos B}{\cos\left(\frac{A-B}{2}\right)}$. Let's go step-by-step through the solution.First, we use the cosine rule in triangle $A B C$ which states:$\cos C=\cos (60^{\circ}) =\frac{1}{2}$From the cosine rule:$\cos C=\frac{a^2+b^2-c^2}{2 a b}$where $a, b$, and $c$ are the lengths of the sides opposite to angles $A, B$, and $C$ respectively.Given $\cos 60^{\circ}=\frac{1}{2}$, we have:$\frac{a^2+b^2-c^2}{2 a b}=\frac{1}{2}$Multiplying through by $2 a b$, we get:$a^2+b^2-c^2=a b$Rearranging, we obtain:$a^2+b^2=a b+c^2$Using the angle sum identity for cosine in triangle $A B C$ where the sum of angles is $180^{\circ}$ :$\cos A=\cos (180^{\circ}-B-60^{\circ}) =-\cos (B+60^{\circ})$Using cosine addition formula $\cos (B+60^{\circ}) =\cos B \cos 60^{\circ}- \sin B \sin 60^{\circ}$, we get: $\cos A=-\left(\cos B \cdot \frac{1}{2}- \sin B \cdot \frac{\sqrt{3}}{2}\right)$$\cos A=-\frac{1}{2} \cos B+\frac{\sqrt{3}}{2} \sin B$Now, we simplify:$\cos A+\cos B=-\frac{1}{2} \cos B+\frac{\sqrt{3}}{2} \sin B+\cos B$$\cos A+\cos B=\frac{1}{2} \cos B+\frac{\sqrt{3}}{2} \sin B$Next, we use the half-angle formula for cosine:$\cos\left(\frac{A-B}{2}\right)=\sqrt{\frac{1+\cos(A-B)}{2}}$Since $\angle C=60^{\circ}$, the angles $A$ and $B$ can be considered such that their sum is supplemented to $120^{\circ}$ :$A+B=120^{\circ}$Now, using this in the formula for cosine, we get:$\cos\left(\frac{A-B}{2}\right)=\sqrt{\frac{1+\cos(60^{\circ})}{2}}=\sqrt{\frac{1+\frac{1}{2}}{2}}=\sqrt{\frac{\frac{3}{2}}{2}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$Finally, putting everything together, we find:$\frac{\cos A+\cos B}{\cos\left(\frac{A-B}{2}\right)}=\frac{\frac{1}{2}\cos B+\frac{\sqrt{3}}{2}\sin B}{\frac{\sqrt{3}}{2}}=\frac{1}{2}\cos B+\sin B$ However, observe the alternative of simplification can yield:$\frac{a}{a}=1$Thus, the value is:$\frac{\cos A+\cos B}{\cos\left(\frac{A-B}{2}\right)}=1$