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प्रश्न
If sin θ = `1/2`, then find the value of θ.
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उत्तर
sin θ = `1/2`
`sin 30^circ = 1/2` ................ [using trignometric table]
∴ θ = 30°
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संबंधित प्रश्न
Prove the identity (sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θ.
Prove the following trigonometric identities.
`(1 + cos θ + sin θ)/(1 + cos θ - sin θ) = (1 + sin θ)/cos θ`
Prove the following trigonometric identities.
`(1 + cos theta - sin^2 theta)/(sin theta (1 + cos theta)) = cot theta`
Prove the following identities:
(1 + cot A – cosec A)(1 + tan A + sec A) = 2
Prove the following identities:
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (1 + cosA)/sinA`
`(1+ tan^2 theta)/(1+ tan^2 theta)= (cos^2 theta - sin^2 theta)`
If cosec2 θ (1 + cos θ) (1 − cos θ) = λ, then find the value of λ.
Write True' or False' and justify your answer the following :
The value of \[\sin \theta\] is \[x + \frac{1}{x}\] where 'x' is a positive real number .
\[\frac{x^2 - 1}{2x}\] is equal to
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Prove the following identity :
`(1 - sin^2θ)sec^2θ = 1`
Prove the following identity :
`(sec^2θ - sin^2θ)/tan^2θ = cosec^2θ - cos^2θ`
Without using trigonometric identity , show that :
`sin42^circ sec48^circ + cos42^circ cosec48^circ = 2`
Evaluate:
sin2 34° + sin2 56° + 2 tan 18° tan 72° – cot2 30°
Prove that `(sec θ - 1)/(sec θ + 1) = ((sin θ)/(1 + cos θ ))^2`
Prove that `((tan 20°)/(cosec 70°))^2 + ((cot 20°)/(sec 70°))^2 = 1`
Prove the following identities.
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
If tan θ × A = sin θ, then A = ?
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
If 4 tanβ = 3, then `(4sinbeta-3cosbeta)/(4sinbeta+3cosbeta)=` ______.
