Advertisements
Advertisements
प्रश्न
Prove that sec2θ − cos2θ = tan2θ + sin2θ
उत्तर
L.H.S = sec2θ − cos2θ
= 1 + tan2θ – cos2θ .......[∵ 1 + tan2θ = sec2θ]
= tan2θ + (1 – cos2θ)
= tan2θ + sin2θ ......`[(because sin^2theta +cos^2theta = 1),(therefore 1 - cos^2theta = sin^2theta)]`
= R.H.S
∴ sec2θ − cos2θ = tan2θ + sin2θ
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`tan theta + 1/tan theta = sec theta cosec theta`
Prove the following trigonometric identities.
sec6θ = tan6θ + 3 tan2θ sec2θ + 1
Prove the following trigonometric identities.
`(cot^2 A(sec A - 1))/(1 + sin A) = sec^2 A ((1 - sin A)/(1 + sec A))`
Prove the following trigonometric identities.
`(1 + cot A + tan A)(sin A - cos A) = sec A/(cosec^2 A) - (cosec A)/sec^2 A = sin A tan A - cos A cot A`
Prove that
`sqrt((1 + sin θ)/(1 - sin θ)) + sqrt((1 - sin θ)/(1 + sin θ)) = 2 sec θ`
Prove that:
`tanA/(1 - cotA) + cotA/(1 - tanA) = secA cosecA + 1`
`(cos theta cosec theta - sin theta sec theta )/(costheta + sin theta) = cosec theta - sec theta`
If `(cot theta ) = m and ( sec theta - cos theta) = n " prove that " (m^2 n)(2/3) - (mn^2)(2/3)=1`
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
Prove the following identity :
`tanA - cotA = (1 - 2cos^2A)/(sinAcosA)`
Prove the following identity :
`cosA/(1 - tanA) + sin^2A/(sinA - cosA) = cosA + sinA`
Prove that sec2 (90° - θ) + tan2 (90° - θ) = 1 + 2 cot2 θ.
Without using a trigonometric table, prove that
`(cos 70°)/(sin 20°) + (cos 59°)/(sin 31°) - 8sin^2 30° = 0`.
Prove the following identities.
(sin θ + sec θ)2 + (cos θ + cosec θ)2 = 1 + (sec θ + cosec θ)2
If 5x = sec θ and `5/x` = tan θ, then `x^2 - 1/x^2` is equal to
If a cos θ – b sin θ = c, then prove that (a sin θ + b cos θ) = `± sqrt("a"^2 + "b"^2 -"c"^2)`
If tan θ + cot θ = 2, then tan2θ + cot2θ = ?
Prove that sin4A – cos4A = 1 – 2cos2A
Prove that `(1 + sin "B")/"cos B" + "cos B"/(1 + sin "B")` = 2 sec B
If cos A + cos2A = 1, then sin2A + sin4 A = ?
