SSC CHSL 2021 PYQ — Question 53

The correct answer is option 4 i.e. 1

  cos$\theta$ = $\frac{2}{3}$

⇒B = 2, H = 3

By Pythagoras theorem, 3² = P² + 2²

P² = 9 - 4

So, P = $\sqrt5$

sec²θ = (3/2)² = 9/4​

2sec²θ = 2 × 9/4 ​ =9/2​

tanθ = $\sqrt5$/2

tan²θ = ($\sqrt5$/2)² = 5/4​

2tan²θ = 5/2

2sec²θ + 2tan²θ − 6 = 9/2 + 5/2 ​− 6 = 1