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प्रश्न
Prove the following identity :
( 1 + cotθ - cosecθ) ( 1 + tanθ + secθ)
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उत्तर
(1 + cotθ - cosecθ) ( 1 + tanθ + secθ)
= `(1 + sinθ/cosθ + 1/cosθ)(1 + cosθ/sinθ - 1/sinθ)`
= `((cosθ + sinθ + 1)/cosθ)((sinθ + cosθ - 1)/sinθ)`
= `((sinθ + cosθ)^2 - (1)^2)/(sinθcosθ)`
= `(sin^2θ + cos^2θ + 2sinθ cosθ - 1)/(sinθcosθ)`
= `(1 + 2sinθ cosθ - 1)/(sinθcosθ)`
= `(2sinθ cosθ)/(sinθ cosθ) = 2`
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संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`sqrt((1+sinA)/(1-sinA)) = secA + tanA`
Prove the following trigonometric identities. `(1 - cos A)/(1 + cos A) = (cot A - cosec A)^2`
Prove the following trigonometric identities.
`sin A/(sec A + tan A - 1) + cos A/(cosec A + cot A + 1) = 1`
Prove the following identities:
`(sinAtanA)/(1 - cosA) = 1 + secA`
Prove that:
`"tanθ"/("secθ" – 1) = (tanθ + secθ + 1)/(tanθ + secθ - 1)`
From the figure find the value of sinθ.
Prove the following identity :
`cosA/(1 - tanA) + sinA/(1 - cotA) = sinA + cosA`
Verify that the points A(–2, 2), B(2, 2) and C(2, 7) are the vertices of a right-angled triangle.
If x sin3 θ + y cos3 θ = sin θ cos θ and x sin θ = y cos θ, then prove that x2 + y2 = 1
sin4A – cos4A = 1 – 2cos2A. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= (sin2A + cos2A) `(square)`
= `1 (square)` ...`[sin^2"A" + square = 1]`
= `square` – cos2A ...[sin2A = 1 – cos2A]
= `square`
= R.H.S.
