SSC CHSL 2024 PYQ — Question 62

The correct answer is option 3 i.e. $-{10\over\sqrt{119}}$.

cot A = 5/12 = Base/Height

By Pythagorean theorem, the hypotenuse is √(5² + 12²) = √(25 + 144) = √169 = 13

sec A = Hypotenuse/Base = 13/5

cosec A = Hypotenuse/Height = 13/12

Now, ${\sqrt{secA-cosecA} \over{\sqrt{secA+cosecA}}} - \sqrt{17\over7}$

$={\sqrt{\frac{13}{5}-\frac{13}{12}} \over{\sqrt{\frac{13}{5}+\frac{13}{12}}}} - \sqrt{17\over7}$

$={\sqrt{\frac{156-65}{60}} \over{\sqrt{\frac{156+65}{60}}}} - \sqrt{17\over7}$

$={\sqrt{\frac{91}{60}} \over{\sqrt{\frac{221}{60}}}} - \sqrt{17\over7}$

$=\sqrt{\frac{91}{221}} - \sqrt{17\over7}$

$=\sqrt{\frac{7}{17}} - \sqrt{\frac{17}{7}}$

$=\frac{\sqrt{7}}{\sqrt{17}} \times \frac{\sqrt{17}}{\sqrt{17}} - \sqrt{\frac{17}{7}}$

$=\frac{7\sqrt{119} - 17\sqrt{119}}{119}$

$=\frac{-10\sqrt{119}}{119}$

$=-{10\over\sqrt{119}}$

Therefore, the value of the given expression is $-{10\over\sqrt{119}}$.