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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`
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Solution
\[\boxed{\text{sec}^2θ} = 1 + \text{tan}^2θ\] ...[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
\[(\text{sec} θ - \text{tan} θ) = \boxed{\frac{1}{\sqrt3}}\]
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