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प्रश्न
If cos A = `(2sqrt(m))/(m + 1)`, then prove that cosec A = `(m + 1)/(m - 1)`.
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उत्तर
`cos A = (2sqrt(m))/(m + 1)` ...[Given]
We know that,
sin2A + cos2A = 1
∴ `sin^2A + ((2sqrt(m))/(m + 1))^2 = 1`
∴ `sin^2A + (4m)/(m + 1)^2 = 1`
∴ `sin^2A = 1 - (4m)/(m + 1)^2`
= `((m + 1)^2 - 4m)/(m + 1)^2`
= `(m^2 + 2m + 1 - 4m)/(m + 1)^2` ...[∵ (a + b)2 = a2 + 2ab + b2]
= `(m^2 - 2m + 1)/(m + 1)^2`
∴ `sin^2A = (m - 1)^2/(m + 1)^2` ...[∵ a2 – 2ab + b2 = (a – b)2]
∴ `sin A = (m - 1)/(m + 1)` ...[Taking square root of both sides]
Now, `"cosec" A = 1/(sin A)`
= `1/((m - 1)/(m + 1))`
∴ `"cosec" A = (m + 1)/(m - 1)`
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संबंधित प्रश्न
Prove that:
sec2θ + cosec2θ = sec2θ x cosec2θ
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 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`
Prove the following trigonometric identities.
`((1 + tan^2 theta)cot theta)/(cosec^2 theta) = tan theta`
Prove the following trigonometric identities.
sin2 A cos2 B − cos2 A sin2 B = sin2 A − sin2 B
If a cos `theta + b sin theta = m and a sin theta - b cos theta = n , "prove that "( m^2 + n^2 ) = ( a^2 + b^2 )`
If tan A = n tan B and sin A = m sin B , prove that `cos^2 A = ((m^2-1))/((n^2 - 1))`
If cosec2 θ (1 + cos θ) (1 − cos θ) = λ, then find the value of λ.
Prove that the following identities:
Sec A( 1 + sin A)( sec A - tan A) = 1.
If sin θ (1 + sin2 θ) = cos2 θ, then prove that cos6 θ – 4 cos4 θ + 8 cos2 θ = 4
If `tan θ = 9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ...[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
Prove that `(sin θ + tan θ)/(cos θ) = tan θ (1 + sec θ)`.
Prove that `(cosθ)/(1 + sinθ) = (1 - sinθ)/(cosθ)`.
Prove that `(1 + sec A)/(sec A) = (sin^2A)/(1 - cos A)`.
Prove that `(cot A + "cosec" A - 1)/(cot A - "cosec" A + 1) = (1 + cos A)/(sin A)`.
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.
Show that: `tan "A"/(1 + tan^2 "A")^2 + cot "A"/(1 + cot^2 "A")^2 = sin"A" xx cos"A"`
Eliminate θ if x = r cosθ and y = r sinθ.
Find the value of sin2θ + cos2θ
Solution:
In Δ ABC, ∠ABC = 90°, ∠C = θ°
AB2 + BC2 = `square` .....(Pythagoras theorem)
Divide both sides by AC2
`"AB"^2/"AC"^2 + "BC"^2/"AC"^2 = "AC"^2/"AC"^2`
∴ `("AB"^2/"AC"^2) + ("BC"^2/"AC"^2) = 1`
But `"AB"/"AC" = square and "BC"/"AC" = square`
∴ `sin^2 theta + cos^2 theta = square`
