Advertisements
Advertisements
प्रश्न
If 2sin2θ – cos2θ = 2, then find the value of θ.
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°
APPEARS IN
संबंधित प्रश्न
Prove the following identities, where the angles involved are acute angles for which the expressions are defined:
`(cosec θ – cot θ)^2 = (1-cos theta)/(1 + cos theta)`
Prove the following trigonometric identities
`((1 + sin theta)^2 + (1 + sin theta)^2)/(2cos^2 theta) = (1 + sin^2 theta)/(1 - sin^2 theta)`
Prove the following identities:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
Prove the following identities:
`1/(cosA + sinA) + 1/(cosA - sinA) = (2cosA)/(2cos^2A - 1)`
`(1-tan^2 theta)/(cot^2-1) = tan^2 theta`
Prove that `( sintheta - 2 sin ^3 theta ) = ( 2 cos ^3 theta - cos theta) tan theta`
Write the value of `(1 + cot^2 theta ) sin^2 theta`.
Write the value of ` sec^2 theta ( 1+ sintheta )(1- sintheta).`
If tanθ `= 3/4` then find the value of secθ.
Prove the following identity :
`sec^2A.cosec^2A = tan^2A + cot^2A + 2`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`tan35^circ cot(90^circ - θ) = 1`
Prove that sin2 θ + cos4 θ = cos2 θ + sin4 θ.
Prove the following identities.
`sqrt((1 + sin theta)/(1 - sin theta)` = sec θ + tan θ
`(1 - tan^2 45^circ)/(1 + tan^2 45^circ)` = ?
Prove that `costheta/(1 + sintheta) = (1 - sintheta)/(costheta)`
Prove that `sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ
Prove that sec2θ – cos2θ = tan2θ + sin2θ
If sinθ = `11/61`, then find the value of cosθ using the trigonometric identity.
Show that, cotθ + tanθ = cosecθ × secθ
Solution :
L.H.S. = cotθ + tanθ
= `cosθ/sinθ + sinθ/cosθ`
= `(square + square)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ............... `square`
= `1/sinθ xx 1/square`
= cosecθ × secθ
L.H.S. = R.H.S
∴ cotθ + tanθ = cosecθ × secθ
