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प्रश्न
Prove the following identities:
cosec4 A (1 – cos4 A) – 2 cot2 A = 1
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उत्तर
cosec4 A (1 – cos4 A) – 2 cot2 A
= cosec4 A (1 – cos2 A) (1 + cos2 A) – 2 cot2 A
= cosec4 A (sin2 A) (1 + cos2 A) – 2 cot2 A
= cosec2 A (1 + cos2 A) – 2 cot2 A
= `cosec^2A + cos^2A/sin^2A - 2cot^2A `
= cosec2 A + cot2 A – 2 cot2 A
= cosec2 A – cot2 A
= 1
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संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 + cos θ + sin θ)/(1 + cos θ - sin θ) = (1 + sin θ)/cos θ`
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If `(cot theta ) = m and ( sec theta - cos theta) = n " prove that " (m^2 n)(2/3) - (mn^2)(2/3)=1`
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`"tanθ"/("secθ" – 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
Without using trigonometric table , evaluate :
`sin72^circ/cos18^circ - sec32^circ/(cosec58^circ)`
Without using trigonometric identity , show that :
`sin42^circ sec48^circ + cos42^circ cosec48^circ = 2`
Prove that:
`(cot A - 1)/(2 - sec^2 A) = cot A/(1 + tan A)`
Prove the following identities.
`(1 - tan^2theta)/(cot^2 theta - 1)` = tan2 θ
Prove that `(tan^2 theta - 1)/(tan^2 theta + 1)` = 1 – 2 cos2θ
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.
