SSC CHSL 2021 PYQ — Question 51

The correct answer is Option 2 i.e. ${{1+P^2} \over 2P}$

Given:

  cosθ  $={{p^2-1} \over p^2+1}$

Formula USed:

cosθ = $Base \over Hypotenuse$

Pythagoras theorem, H² = P² + B²

cosecθ = $Hypotenuse \over Perpendicular$

Explanation:

Accordin to the given question,

B = p² - 1 and H = p² + 1

So, P² = H² - B² ⇒ P² = (p² + 1)² - (p² - 1)²

⇒ P² = 4p²

⇒ P = 2p

Now putting this in the formula of cosecθ, we get

cosecθ = ${p^2 + 1} \over 2p$

Hence the required answer is ${p^2 + 1} \over 2p$ .