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Question
If `secθ = 25/7 ` then find tanθ.
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Solution
`1 + tan^2θ = sec^2θ`
`1 + tan^2θ =(25/7)^2`
`∴ tan^2θ =625/49- 1`
`∴ tan^2θ =(625-49)/49`
`∴ tan^2θ =576/49`
`∴ tanθ =24/7`
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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)
Choose the correct alternative:
`(1 + cot^2"A")/(1 + tan^2"A")` = ?
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Activity:
`5/(sin^2theta) - 5cot^2theta`
= `square (1/(sin^2theta) - cot^2theta)`
= `5(square - cot^2theta) ......[1/(sin^2theta) = square]`
= 5(1)
= `square`
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