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प्रश्न
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
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उत्तर
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
LHS = `(cotA + cosecA - 1)/(cotA - cosecA + 1)`
we know that , cosec2A - cot2A = 1
substituting this in the numerator
`(cosecA + cotA - (cosec^2A - cot^2A))/(cotA - cosecA + 1)` .....(x²-y²= (x+y)(x-y))
`(cosecA + cotA - (cosecA + cotA)(cosecA - cotA))/(cotA - cosecA + 1)`
taking common
`((cosec A + cot A)(1-cosec A + cot A) )/ (cot A - cosec A + 1)`
cancelling like terms in numerator and denominator
we are left with cosec A + cot A
`= 1/sin A + cos A/sin A`
`= (1+cos A) / sin A`
= RHS
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संबंधित प्रश्न
Prove the following trigonometric identities.
`1/(1 + sin A) + 1/(1 - sin A) = 2sec^2 A`
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`(tan^2 A)/(1 + tan^2 A) + (cot^2 A)/(1 + cot^2 A) = 1`
Prove the following identities:
(1 + tan A + sec A) (1 + cot A – cosec A) = 2
Prove the following identity :
`sinA/(1 + cosA) + (1 + cosA)/sinA = 2cosecA`
Prove the following identity :
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
Prove that: `sqrt((1 - cos θ)/(1 + cos θ)) = cosec θ - cot θ`.
Prove that the following identities:
Sec A( 1 + sin A)( sec A - tan A) = 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
Proved that `(1 + secA)/secA = (sin^2A)/(1 - cos A)`.
Prove the following trigonometry identity:
(sin θ + cos θ)(cosec θ – sec θ) = cosec θ ⋅ sec θ – 2 tan θ
