Advertisements
Advertisements
प्रश्न
Prove that `(sin^2θ)/(cos θ) + cos θ = sec θ`.
Advertisements
उत्तर
L.H.S. = `(sin^2θ)/(cos θ) + cos θ`
= `(sin^2θ + cos^2θ)/(cos θ)`
= `1/(cos θ)` ...[∵ sin2θ + cos2θ = 1]
= sec θ
= R.H.S.
∴ `(sin^2θ)/(cos θ) + cos θ = sec θ`
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`(1 + cos theta - sin^2 theta)/(sin theta (1 + cos theta)) = cot theta`
Prove the following identities:
sec4 A (1 – sin4 A) – 2 tan2 A = 1
Prove the following identities:
cosec4 A (1 – cos4 A) – 2 cot2 A = 1
`sin^6 theta + cos^6 theta =1 -3 sin^2 theta cos^2 theta`
`(sectheta- tan theta)/(sec theta + tan theta) = ( cos ^2 theta)/( (1+ sin theta)^2)`
`(tan A + tanB )/(cot A + cot B) = tan A tan B`
Prove that `( sintheta - 2 sin ^3 theta ) = ( 2 cos ^3 theta - cos theta) tan theta`
Write the value of`(tan^2 theta - sec^2 theta)/(cot^2 theta - cosec^2 theta)`
Prove that:
`(sin^2θ)/(cosθ) + cosθ = secθ`
If x = r sin θ cos ϕ, y = r sin θ sin ϕ and z = r cos θ, then
Prove the following identity :
`(1 - sin^2θ)sec^2θ = 1`
Prove the following identity :
`tanA - cotA = (1 - 2cos^2A)/(sinAcosA)`
If m = a secA + b tanA and n = a tanA + b secA , prove that m2 - n2 = a2 - b2
Prove that:
`(cot A - 1)/(2 - sec^2 A) = cot A/(1 + tan A)`
Prove that sin θ sin( 90° - θ) - cos θ cos( 90° - θ) = 0
Prove that `( tan A + sec A - 1)/(tan A - sec A + 1) = (1 + sin A)/cos A`.
Prove that: `1/(sec θ - tan θ) = sec θ + tan θ`.
`sqrt((1 - cos^2theta) sec^2 theta) = tan theta`
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
If sin A = `1/2`, then the value of sec A is ______.
