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प्रश्न
tan θ × `sqrt(1 - sin^2 θ)` is equal to:
पर्याय
cos θ
sin θ
tan θ
cot θ
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उत्तर
sin θ
Explanation:
`tan θ xx sqrt(1 - sin^2 θ) ...{sin^2 θ + cos^2 θ = 1, ∴ cos^2 θ = 1 - sin^2 θ}`
= `tan θ xx sqrt(cos^2 θ)`
= tan θ × cos θ
= `(sin θ)/(cos θ)` × cos θ
= sin θ
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संबंधित प्रश्न
Prove the following trigonometric identities.
`(tan^2 A)/(1 + tan^2 A) + (cot^2 A)/(1 + cot^2 A) = 1`
Prove the following identities:
`cot^2A/(cosecA + 1)^2 = (1 - sinA)/(1 + sinA)`
Prove the following identities:
`(sinAtanA)/(1 - cosA) = 1 + secA`
If m = a sec A + b tan A and n = a tan A + b sec A, then prove that : m2 – n2 = a2 – b2
Prove that:
(cosec θ - sinθ )(secθ - cosθ ) ( tanθ +cot θ) =1
Without using trigonometric table , evaluate :
`sin72^circ/cos18^circ - sec32^circ/(cosec58^circ)`
If A + B = 90°, show that sec2 A + sec2 B = sec2 A. sec2 B.
Prove that: `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(sin^2 A - cos^2 A)`.
Prove that `"cosec" θ xx sqrt(1 - cos^2θ) = 1`.
If `sec θ + tan θ = sqrt(3)`, complete the activity to find the value of sec θ – tan θ.
Activity:
`square = 1 + tan^2θ` ...[Fundamental trigonometric identity]
`square - tan^2θ = 1`
`(sec θ + tan θ) . (sec θ - tan θ) = square`
`sqrt(3) . (sec θ - tan θ) = 1`
`(sec θ - tan θ) = square`
