Question

If cosec θ = `sqrt5`, find the value of:

1. 2 - sin² θ - cos² θ
2. 2 + `1/sin^2"θ" – cos^2"θ"/sin^2"θ"` 

Answer 1

Consider the diagram below :

cosec θ =`sqrt5`

i.e.`"hypotenuse"/"perpendicular" = sqrt5/1`

Therefore, if length of hypotenuse = `sqrt5`, length of perpendicular = x

Since
base² + perpendicular² = hypotenuse²              ...[Using Pythagoras Theorem]

base² + (x)² = `(sqrt5x)^2`

base² = 5x² – x²

base² = 4x²

∴ base = 2x

Now

sin θ = `"perpendicular"/"hypotenuse" = (x)/(sqrt5x) = (1)/(sqrt5)`

cos θ = `"base"/"hypotenuse" = (2)/(sqrt5x) = (2)/(sqrt5)`

(i) 2 – sin² θ – cos² θ

= 2 – `(1/sqrt5)^2 – (2/sqrt5)^2`

= 2 – `(1)/(5) –(4)/(5)`

= `(5)/(5)`

= 1

(ii) 2 + `1/sin^2"θ"  –  cos^2"θ"/sin^2"θ"` 

= 2 + `1/(x/sqrt(5x))^2 – ((2x)/(sqrt5x))^2/(x/(sqrt5)^2)`

= `2 + 5  –  (4/5)/(1/5)`

= 2 + 5 – 4

= 7 – 4

= 3