Advertisements
Advertisements
प्रश्न
Prove that sec2θ – cos2θ = tan2θ + sin2θ.
Advertisements
उत्तर
L.H.S. = sec2θ – cos2θ
= sec2θ – (1 – sin2θ) ...`[(∵ sin^2θ + cos^2θ = 1),(∴ 1 - sin^2θ = cos^2θ)]`
= sec2θ – 1 + sin2θ
= tan2θ + sin2θ ...`[(∵ 1 + tan^2θ = sec^2θ),(∴ tan^2θ = sec^2θ - 1)]`
= R.H.S.
∴ sec2θ – cos2θ = tan2θ + sin2θ
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`"cosec" theta sqrt(1 - cos^2 theta) = 1`
Prove the following identities:
`(sintheta - 2sin^3theta)/(2cos^3theta - costheta) = tantheta`
Prove the following identities:
`(costhetacottheta)/(1 + sintheta) = cosectheta - 1`
Prove that:
`1/(cosA + sinA - 1) + 1/(cosA + sinA + 1) = cosecA + secA`
Show that : tan 10° tan 15° tan 75° tan 80° = 1
Write the value of ` cosec^2 (90°- theta ) - tan^2 theta`
Write the value of `3 cot^2 theta - 3 cosec^2 theta.`
If sec θ + tan θ = x, write the value of sec θ − tan θ in terms of x.
If sec2 θ (1 + sin θ) (1 − sin θ) = k, then find the value of k.
If cos (\[\alpha + \beta\]= 0 , then sin \[\left( \alpha - \beta \right)\] can be reduced to
Prove the following identity :
`(1 - tanA)^2 + (1 + tanA)^2 = 2sec^2A`
Prove the following identity :
`((1 + tan^2A)cotA)/(cosec^2A) = tanA`
Prove that `(cos θ)/(1 - sin θ) = (1 + sin θ)/(cos θ)`.
Without using the trigonometric table, prove that
cos 1°cos 2°cos 3° ....cos 180° = 0.
Prove the following identities.
`sqrt((1 + sin theta)/(1 - sin theta)) + sqrt((1 - sin theta)/(1 + sin theta))` = 2 sec θ
Prove that `(sin θ + "cosec" θ)/(sin θ) = 2 + cot^2θ`.
Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.
Simplify (1 + tan2θ)(1 – sinθ)(1 + sinθ)
If tan θ + sec θ = l, then prove that sec θ = `(l^2 + 1)/(2l)`.
If sinθ = `11/61`, then find the value of cosθ using the trigonometric identity.
