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प्रश्न
If tan θ = `9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ......[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
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उत्तर
sec2θ = 1 + tan2θ ......[Fundamental trigonometric identity]
∴ sec2θ = 1 + `(9/40)^2`
∴ sec2θ = 1 + `81/1600`
∴ sec2θ = `1681/1600`
∴ sec θ = `41/40`
संबंधित प्रश्न
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Prove that: `1/(sec θ - tan θ) = sec θ + tan θ`.
If 1 + sin2θ = 3 sin θ cos θ, then prove that tan θ = 1 or `1/2`.
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
Prove the following identity:
(sin2θ – 1)(tan2θ + 1) + 1 = 0
