UPSC NDA II 2015 Mathematics Paper

Concept:Trigonometric functions have periodic properties. For any integer $n$, $\tan(2n\pi \pm \theta) = \tan(\pm \theta)$ and $\cot(2n\pi \pm \theta) = \cot(\pm \theta)$. Also, $\cot(π + \theta) = \cot \theta$. We reduce given angles to standard angles between $0$ and $2π$.Explanation:Compute $p = \tan\left(-\frac{11\pi}{6}\right)$. $-\frac{11\pi}{6} = -2\pi + \frac{\pi}{6}$. Thus $p = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$. Compute $q = \tan\left(\frac{21\pi}{4}\right)$. $\frac{21\pi}{4} = 4\pi + \frac{5\pi}{4}$. Now $\frac{5\pi}{4} = \pi + \frac{\pi}{4}$, so $\tan\left(\frac{5\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) = 1$. Hence $q = 1$. Compute $r = \cot\left(\frac{283\pi}{6}\right)$. $\frac{283\pi}{6} = 46\pi + \frac{7\pi}{6}$. Since $\frac{7\pi}{6} = \pi + \frac{\pi}{6}$, we have $\cot\left(\frac{7\pi}{6}\right) = \cot\left(\frac{\pi}{6}\right) = \sqrt{3}$. Thus $r = \sqrt{3}$. Now evaluate statement 1: $p \times r = \frac{1}{\sqrt{3}} \times \sqrt{3} = 1$, not $2$. So statement 1 is false. Evaluate statement 2: Check if $p, q, r$ are in geometric progression. $p = \frac{1}{\sqrt{3}}$, $q = 1$, $r = \sqrt{3}$. Common ratio: $\frac{q}{p} = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3}$, and $\frac{r}{q} = \frac{\sqrt{3}}{1} = \sqrt{3}$. Since the ratio is constant, $p, q, r$ are in GP. Statement 2 is true.Answer:Only statement 2 is correct. Hence the answer is option B (2 only).