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प्रश्न
If cosec θ = 2x and \[5\left( x^2 - \frac{1}{x^2} \right)\] \[2\left( x^2 - \frac{1}{x^2} \right)\]
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उत्तर
Given:
`cosecθ=2x, cot θ2/x`
We know that,
`cosec^2 θ-cot^2 θ=1`
⇒` (2x)^2-(2/x)^2=1`
⇒` 4x^2-4/x^2=1`
⇒ `4(x^2-1/x^2)=1`
⇒`2xx2xx(x^2-1/x^2)=1`
⇒ `2(x^2-1/x^2)=1/2`
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संबंधित प्रश्न
9 sec2 A − 9 tan2 A = ______.
Prove the following trigonometric identities
(1 + cot2 A) sin2 A = 1
Prove the following trigonometric identities.
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Prove the following trigonometric identities.
`(cos theta - sin theta + 1)/(cos theta + sin theta - 1) = cosec theta + cot theta`
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Prove the following identities:
`cosecA + cotA = 1/(cosecA - cotA)`
Prove the following identities:
`cot^2A/(cosecA + 1)^2 = (1 - sinA)/(1 + sinA)`
Prove the following identities:
`sqrt((1 + sinA)/(1 - sinA)) = cosA/(1 - sinA)`
Prove the following identities:
sec4 A (1 – sin4 A) – 2 tan2 A = 1
Prove that:
cos A (1 + cot A) + sin A (1 + tan A) = sec A + cosec A
`1+ (cot^2 theta)/((1+ cosec theta))= cosec theta`
If `tan theta = 1/sqrt(5), "write the value of" (( cosec^2 theta - sec^2 theta))/(( cosec^2 theta - sec^2 theta))`.
Prove the following identity :
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Prove the following identity :
`sin^4A + cos^4A = 1 - 2sin^2Acos^2A`
Prove the following identity :
`tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ) = 1/(sin^2θ - cos^2θ)`
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`sin72^circ/cos18^circ - sec32^circ/(cosec58^circ)`
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sin2θ + sin2(90 – θ) = ?
Prove that cot2θ × sec2θ = cot2θ + 1
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`
