Advertisements
Advertisements
प्रश्न
Prove that sec2θ – cos2θ = tan2θ + sin2θ.
Advertisements
उत्तर
L.H.S. = sec2θ – cos2θ
= 1 + tan2θ – cos2θ ...[∵ 1 + tan2θ = sec2θ]
= tan2θ + (1 – cos2θ)
= tan2θ + sin2θ ...`[(∵ sin^2θ +cos^2θ = 1),(∴ 1 - cos^2θ = sin^2θ)]`
= R.H.S.
∴ sec2θ – cos2θ = tan2θ + sin2θ
APPEARS IN
संबंधित प्रश्न
Evaluate
`(sin ^2 63^@ + sin^2 27^@)/(cos^2 17^@+cos^2 73^@)`
9 sec2 A − 9 tan2 A = ______.
Prove the following trigonometric identities.
`cot theta - tan theta = (2 cos^2 theta - 1)/(sin theta cos theta)`
Prove the following trigonometric identities.
(sec A − cosec A) (1 + tan A + cot A) = tan A sec A − cot A cosec A
Prove the following trigonometric identities.
`(tan A + tan B)/(cot A + cot B) = tan A tan B`
If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan θ, show that `x^2/a^2 + y^2/b^2 - z^2/c^2 = 1`
If x = r sin A cos B, y = r sin A sin B and z = r cos A, then prove that : x2 + y2 + z2 = r2
`(1 + cot^2 theta ) sin^2 theta =1`
If 3 `cot theta = 4 , "write the value of" ((2 cos theta - sin theta))/(( 4 cos theta - sin theta))`
If cosec θ = 2x and \[5\left( x^2 - \frac{1}{x^2} \right)\] \[2\left( x^2 - \frac{1}{x^2} \right)\]
Prove the following identity :
( 1 + cotθ - cosecθ) ( 1 + tanθ + secθ)
Prove the following identity :
`(tanθ + 1/cosθ)^2 + (tanθ - 1/cosθ)^2 = 2((1 + sin^2θ)/(1 - sin^2θ))`
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
Prove that:
`(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(2 sin^2 A - 1)`
If (sin α + cosec α)2 + (cos α + sec α)2 = k + tan2α + cot2α, then the value of k is equal to
If 5x = sec θ and `5/x` = tan θ, then `x^2 - 1/x^2` is equal to
Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0
Prove that `(cos^2θ)/(sinθ) + sin θ = "cosec" θ`.
Prove that sec2θ – cos2θ = tan2θ + sin2θ.
If `sqrt(3) tan θ` = 1, then find the value of sin2θ – cos2θ.
