Question

If sec θ + tan θ = `sqrt(3)`, complete the activity to find the value of sec θ – tan θ

Activity:

`square` = 1 + tan²θ    ......[Fundamental trigonometric identity]

`square` – tan²θ = 1

(sec θ + tan θ) . (sec θ – tan θ) = `square`

`sqrt(3)*(sectheta - tan theta)` = 1

(sec θ – tan θ) = `square`

Answer 1

sec²θ = 1 + tan²θ    ......[Fundamental trigonometric identity]

sec²θ – tan²θ = 1

(sec θ + tan θ) . (sec θ – tan θ) = 1

`sqrt(3)*(sectheta - tan theta)` = 1

(sec θ – tan θ) = `1/sqrt(3)`