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प्रश्न
If `1 - cos^2θ = 1/4`, then θ = ?
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उत्तर
`1 - cos^2θ = 1/4` ...[Given]
∴ `sin^2θ = 1/4` ...`[(∵ sin^2θ + cos^2θ = 1),(∴ 1 - cos^2θ = sin^2θ)]`
∴ `sin θ = 1/2` ...[Taking square root of both sides]
∴ θ = 30° ...`[∵ sin 30^circ = 1/2]`
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संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 + sec theta)/sec theta = (sin^2 theta)/(1 - cos theta)`
Prove the following trigonometric identity.
`(sin theta - cos theta + 1)/(sin theta + cos theta - 1) = 1/(sec theta - tan theta)`
Prove the following trigonometric identities.
`1/(sec A + tan A) - 1/cos A = 1/cos A - 1/(sec A - tan A)`
Prove the following trigonometric identities.
`(cot A + tan B)/(cot B + tan A) = cot A tan B`
Prove that:
`(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)`
If sec A + tan A = p, show that:
`sin A = (p^2 - 1)/(p^2 + 1)`
Prove that:
`sqrt(sec^2A + cosec^2A) = tanA + cotA`
`cot^2 theta - 1/(sin^2 theta ) = -1`a
`tan theta/(1+ tan^2 theta)^2 + cottheta/(1+ cot^2 theta)^2 = sin theta cos theta`
` (sin theta - cos theta) / ( sin theta + cos theta ) + ( sin theta + cos theta ) / ( sin theta - cos theta ) = 2/ ((2 sin^2 theta -1))`
If `sin theta = 1/2 , " write the value of" ( 3 cot^2 theta + 3).`
If `cos theta = 2/3 , "write the value of" ((sec theta -1))/((sec theta +1))`
If `sin theta = x , " write the value of cot "theta .`
Prove the following identity :
`(tanθ + 1/cosθ)^2 + (tanθ - 1/cosθ)^2 = 2((1 + sin^2θ)/(1 - sin^2θ))`
For ΔABC , prove that :
`tan ((B + C)/2) = cot "A/2`
Prove that `sin^2 θ/ cos^2 θ + cos^2 θ/sin^2 θ = 1/(sin^2 θ. cos^2 θ) - 2`.
Prove that `cos θ/sin(90° - θ) + sin θ/cos (90° - θ) = 2`.
Prove that: `(sin θ - 2sin^3 θ)/(2 cos^3 θ - cos θ) = tan θ`.
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)
Simplify (1 + tan2θ)(1 – sinθ)(1 + sinθ)
