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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 + \[\boxed{\text{tan}^2θ}\] ...[Fundamental trigonometric identity]
∴ sec2θ = 1 + \[\boxed{\frac{9}{40}}^2\]
∴ sec2θ = 1 + \[\boxed{\frac{81}{1600}}\]
∴ sec2θ = `1681/1600`
∴ sec θ = \[\boxed{\frac{41}{40}}\]
संबंधित प्रश्न
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To prove cot θ + tan θ = cosec θ × sec θ, complete the activity given below.
Activity:
L.H.S. = `square`
= `square/(sinθ) + (sinθ)/(cosθ)`
= `(cos^2θ + sin^2θ)/square`
= `1/(sinθ.cosθ)` ...`[cos^2θ + sin^2θ = square]`
= `1/(sinθ) xx 1/square`
= `square`
= R.H.S.
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