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Question
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
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Solution
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^2`
∴ tan2 θ = `625/49 - 1`
= `(625 - 49)/49`
= `576/49`
∴ tan θ = `24/7` ........(by taking square roots)
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