SSC CGL 2024 PYQ — Question 57

The correct answer is option 3 i.e. 1.

sin²θ + cos²θ = 1

Now, $[1−sin^232°+{1\over 1+tan^258°}]$

$=[cos^232°+{1\over 1+tan^258°}]$   [∵ cos²θ = 1 - sin²θ]

= cos²32° + (1/sec²58°)   [∵ sec²θ = 1 + tan²θ]

= cos²32° + cos²58°   [∵ 1/sec²θ = cos²θ]

= cos²32° + cos²(90° - 32°)

= cos²32° + sin²32°

= 1