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प्रश्न
If 2sin2θ – cos2θ = 2, then find the value of θ.
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उत्तर
Given,
2sin2θ – cos2θ = 2
⇒ 2sin2θ – (1 – sin2θ) = 2 ...[∵ sin2θ + cos2θ = 1]
⇒ 2sin2θ + sin2θ – 1 = 2
⇒ 3sin2θ = 3
⇒ sin2θ = 1
⇒ sinθ = 1 = sin 90° ...[∵ sin 90° = 1]
∴ θ = 90°
संबंधित प्रश्न
Prove the following trigonometric identities.
`sqrt((1 - cos theta)/(1 + cos theta)) = cosec theta - cot theta`
Prove the following trigonometric identities.
`sqrt((1 - cos A)/(1 + cos A)) = cosec A - cot A`
Prove the following trigonometric identities.
sin2 A cos2 B − cos2 A sin2 B = sin2 A − sin2 B
Prove the following identities:
`cot^2A/(cosecA + 1)^2 = (1 - sinA)/(1 + sinA)`
`cot^2 theta - 1/(sin^2 theta ) = -1`a
If `( sin theta + cos theta ) = sqrt(2) , " prove that " cot theta = ( sqrt(2)+1)`.
Write the value of `(cot^2 theta - 1/(sin^2 theta))`.
If a cos θ + b sin θ = m and a sin θ − b cos θ = n, then a2 + b2 =
9 sec2 A − 9 tan2 A is equal to
If tanA + sinA = m and tanA - sinA = n , prove that (`m^2 - n^2)^2` = 16mn
Prove that:
tan (55° + x) = cot (35° – x)
Evaluate:
`(tan 65°)/(cot 25°)`
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
Prove that `sin A/(sec A + tan A - 1) + cos A/("cosec" A + cot A - 1) = 1`.
Prove that `cos θ/sin(90° - θ) + sin θ/cos (90° - θ) = 2`.
Prove that identity:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
Prove that: sin6θ + cos6θ = 1 - 3sin2θ cos2θ.
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
The value of 2sinθ can be `a + 1/a`, where a is a positive number, and a ≠ 1.
`1/sin^2θ - 1/cos^2θ - 1/tan^2θ - 1/cot^2θ - 1/sec^2θ - 1/("cosec"^2θ) = -3`, then find the value of θ.
