UPSC NDA I 2024 Mathematics Paper

Firstly, we need to simplify the expression inside the square root: $\sqrt{15+\cot^2\left(\frac{\pi}{4}-2\cot^{-1}3\right)}$.Let's start by simplifying the term inside the cotangent function: $\frac{\pi}{4}-2\cot^{-1}3$.If we let $x=\cot^{-1}3$, then by definition: $\cot x=3$.Thus, $x=\cot^{-1}3$.Now we need to find: $\cot\left(\frac{\pi}{4}-2x\right)$.Using the cotangent subtraction formula:$\cot(A-B)=\frac{\cot A\cot B+1}{\cot B-\cot A}$Here, $A=\frac{\pi}{4}$ and $B=2x$.Since $\cot\left(\frac{\pi}{4}\right)=1$, we have:$\cot\left(\frac{\pi}{4}-2\cot^{-1}3\right)=\frac{1\cdot\cot(2x)+1}{\cot(2x)-1}$Next, we need to find $\cot(2\cot^{-1}3)$. Using the double-angle formula for cotangent: $\cot(2\theta)=\frac{\cot^2\theta-1}{2\cot\theta}$.In this case, $\theta=\cot^{-1}3$, so $\cot\theta=3$.Then, we get:$\cot(2\theta)=\frac{3^2-1}{2\cdot3}=\frac{9-1}{6}=\frac{8}{6}=\frac{4}{3}\;\text{". "}\;$Therefore, $\cot(2\cot^{-1}3)=\frac{4}{3}$. $\cot\left(\frac{\pi}{4}-2\cot^{-1}3\right)=\frac{1\cdot\frac{4}{3}+1}{\frac{4}{3}-1}=\frac{\frac{4}{3}+1}{\frac{4}{3}-1}=\frac{\frac{4}{3}+\frac{3}{3}}{\frac{4}{3}-\frac{3}{3}}=\frac{\frac{7}{3}}{\frac{1}{3}}=7\;\text{". "}\;$Finally, substituting this result into the original expression gives:$\sqrt{15+\cot^2\left(\frac{\pi}{4}-2\cot^{-1}3\right)}=\sqrt{15+7^2}=\sqrt{15+49}=\sqrt{64}=8\;\text{". "}\;$