SSC CHSL 2023 PYQ — Question 70

The correct answer is Option 4 i.e. tan²A.

$\left (\frac{1+sec^2A}{1+cos^2A} \right ) \left (\frac{1+sin^2A}{1+cosec^2A} \right )$

$=\left( \frac{1 + \frac{1}{\cos^2 A}}{1 + \cos^2 A} \right) \left( \frac{1 + \sin^2 A}{1 + \frac{1}{\sin^2 A}} \right)$

$= \left( \frac{\frac{\cos^2 A + 1}{\cos^2 A}}{1 + \cos^2 A} \right) \left( \frac{1 + \sin^2 A}{\frac{\sin^2 A + 1}{\sin^2 A}} \right)$

$= \left( \frac{\cos^2 A + 1}{\cos^2 A (1 + \cos^2 A)} \right) \left( \frac{(1 + \sin^2 A) \sin^2 A}{\sin^2 A + 1} \right)$

$= \left( \frac{1}{\cos^2 A} \right) \left( \frac{\sin^2 A}{1} \right)$

= tan²A