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प्रश्न
Prove the following identity :
`tanA - cotA = (1 - 2cos^2A)/(sinAcosA)`
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उत्तर
`tanA - cotA = sinA/cosA - cosA/sinA`
= `(sin^2A - cos^2A)/(sinAcosA)`
= `(1 - cos^2A - cos^2A)/(sinAcosA)` (`Q sin^2A = 1 - cos^2A`)
= `(1 - 2cos^2A)/(sinAcosA)`
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संबंधित प्रश्न
Prove the following trigonometric identities.
`"cosec" theta sqrt(1 - cos^2 theta) = 1`
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = cosA/(1 - sinA)`
If `cosec theta = 2x and cot theta = 2/x ," find the value of" 2 ( x^2 - 1/ (x^2))`
If sin2 θ cos2 θ (1 + tan2 θ) (1 + cot2 θ) = λ, then find the value of λ.
If x = a sec θ and y = b tan θ, then b2x2 − a2y2 =
Prove the following identities:
`(1 - tan^2 θ)/(cot^2 θ - 1) = tan^2 θ`.
sec2θ – tan2θ = ?
Prove that cot2θ – tan2θ = cosec2θ – sec2θ.
If cot θ = `40/9`, find the values of cosec θ and sinθ,
We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
`(square + square)/square` = cosec2θ
`square/square` = cosec2θ ......[Taking root on the both side]
cosec θ = `41/9`
and sin θ = `1/("cosec" θ)`
sin θ = `1/square`
∴ sin θ = `9/41`
The value is cosec θ = `41/9`, and sin θ = `9/41`
Prove the following trigonometry identity:
(sin θ + cos θ)(cosec θ – sec θ) = cosec θ ⋅ sec θ – 2 tan θ
