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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 = \[\boxed{\frac{41}{40}}\]
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Prove that:
`(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(2 sin^2 A - 1)`
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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)
1 + cot2θ = ?
If `1 - cos^2θ = 1/4`, then θ = ?
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
(1 – cos2 A) is equal to ______.
