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प्रश्न
Prove the following identity :
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
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उत्तर
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
LHS = `(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1)`
= `(cosec^2A + cosecA + cosec^2A - cosecA)/(cosec^2A - 1)`
= `(2cosec^2A)/cot^2A(Q cosec^2A - 1 = cot^2A)`
= `(2/sin^2A)/(cos^2A/sin^2A) = 2/cos^2A = 2sec^2A`
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संबंधित प्रश्न
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Prove the following trigonometric identities.
`(1 - tan^2 A)/(cot^2 A -1) = tan^2 A`
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`(1+ cos theta - sin^2 theta )/(sin theta (1+ cos theta))= cot theta`
If x=a `cos^3 theta and y = b sin ^3 theta ," prove that " (x/a)^(2/3) + ( y/b)^(2/3) = 1.`
Prove the following identity :
`sinθ(1 + tanθ) + cosθ(1 +cotθ) = secθ + cosecθ`
Prove that: `sqrt((1 - cos θ)/(1 + cos θ)) = "cosec" θ - cot θ`.
tan θ × `sqrt(1 - sin^2 θ)` is equal to:
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
