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प्रश्न
If sec A + tan A = p, show that:
`sin A = (p^2 - 1)/(p^2 + 1)`
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उत्तर
`(p^2 - 1)/(p^2 + 1)`
= `((secA + tanA)^2 - 1)/((secA + tanA)^2 + 1)`
= `(sec^2A + tan^2A + 2tanA secA - 1)/(sec^2A + tan^2A + 2tanA secA + 1)`
= `(tan^2A + tan^2A + 2tanA secA)/(sec^2A + sec^2A + 2tanA secA)`
= `(2tan^2A + 2tanA secA)/(2sec^2A + 2tanA secA)`
= `(2tanA(tanA + secA))/(2secA(tanA + secA)`
= sin A
संबंधित प्रश्न
Prove that `(tan^2 theta)/(sec theta - 1)^2 = (1 + cos theta)/(1 - cos theta)`
Prove the following trigonometric identities.
`(1 - sin θ)/(1 + sin θ) = (sec θ - tan θ)^2`
If `sin theta = 1/2 , " write the value of" ( 3 cot^2 theta + 3).`
`If sin theta = cos( theta - 45° ),where theta " is acute, find the value of "theta` .
Prove the following identity :
cosecθ(1 + cosθ)(cosecθ - cotθ) = 1
Prove the following identity :
`(1 + sinA)/(1 - sinA) = (cosecA + 1)/(cosecA - 1)`
If `asin^2θ + bcos^2θ = c and p sin^2θ + qcos^2θ = r` , prove that (b - c)(r - p) = (c - a)(q - r)
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
Prove the following:
`1 + (cot^2 alpha)/(1 + "cosec" alpha)` = cosec α
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.
