Question

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

Activity:

`square = 1 + tan^2θ`   ...[Fundamental trigonometric identity]

`square - tan^2θ = 1`

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

`sqrt(3)  . (sec θ - tan θ) = 1`

`(sec θ - tan θ) = square`

Answer 1

\[\boxed{\text{sec}^2θ} = 1 + \text{tan}^2θ\]   ...[Fundamental trigonometric identity]

\[\boxed{\text{sec}^2θ} - \text{tan}^2θ = 1\]

(sec θ + tan θ) . (sec θ – tan θ) = \[\boxed{1}\]

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

\[(\text{sec} θ - \text{tan} θ) = \boxed{\frac{1}{\sqrt3}}\]