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प्रश्न
Prove the following trigonometric identities.
`(1 + cos A)/sin A = sin A/(1 - cos A)`
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उत्तर
We need to prove `(1 + cos A)/sin A = sin A/(1 - cos A)`
Now, multiplying the numerator and denominator of LHS by `1 - cos A` we get
`(1 + cos A)/sin A = (1 + cos A)/sin A xx (1 - cos A)/(1 - cos A)`
Further using the identity, `a^2 - b^2 = (a + b)(a - b)` we get
`(1 + cos A)/sin A xx (1 - cos A)/(1 - cos A) = (1 - cos^2 A)/(sin A (1- cos A))`
`= sin^2 A/(sin A(1 - cos A))` (Using `sin^2 theta + cos^2 theta = 1`)
`= sin A/(1 - cos A)`
Hence proved
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संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 + sec theta)/sec theta = (sin^2 theta)/(1 - cos theta)`
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Prove the following trigonometric identities
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Write the value of `( 1- sin ^2 theta ) sec^2 theta.`
If `cot theta = 1/ sqrt(3) , "write the value of" ((1- cos^2 theta))/((2 -sin^2 theta))`
If cosec θ − cot θ = α, write the value of cosec θ + cot α.
Prove the following identity :
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If x = h + a cos θ, y = k + b sin θ.
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Prove the following identities:
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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`
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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.
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.
If sinθ = `11/61`, then find the value of cosθ using the trigonometric identity.
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