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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 + 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`
Solution
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)`
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