SSC CGL 2025 PYQ — Question 54

The correct answer is Option 2 i.e.$15 +8\sqrt3$.

Let t = tan A, then cot A = 1/t

t + 1/t = 4

⇒ t² + 1 = 4t

⇒ t² - 4t + 1 = 0

⇒ $t=\frac{(-4)\pm\sqrt{-(-4)^2-4\times1\times1}}{2\times1}=\frac{-(-4)\pm\sqrt{16-4}}{2}=\frac{4\pm2\sqrt{3}}{2}=2 \pm \sqrt{3}$

⇒ $t^2 = (2 \pm \sqrt{3})^2 = 7 \pm 4\sqrt{3}$

 $\sec^2 A = 1 + \tan^2 A$

⇒ $\tan^2 A + \sec^2 A = \tan^2 A + (1 + \tan^2 A) = 1 + 2\tan^2 A$

$= 1 + 2(7 \pm 4\sqrt{3})$

$= 15 \pm 8\sqrt{3}$