Advertisements
Advertisements
Question
Prove the following identity :
`sec^4A - sec^2A = sin^2A/cos^4A`
Advertisements
Solution
`sec^4A - sec^2A = 1/cos^4A - 1/cos^2A`
= `(1 - cos^2A)/cos^4A`
= `sin^2A/cos^4A` [∵ `sin^2A = 1 - cos^2A`]
APPEARS IN
RELATED QUESTIONS
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
`(sintheta - 2sin^3theta)/(2costheta - costheta) =tan theta`
Prove that `(sin theta)/(1-cottheta) + (cos theta)/(1 - tan theta) = cos theta + sin theta`
Prove the following identities:
`1/(tan A + cot A) = cos A sin A`
Prove the following identities:
`cosA/(1 - sinA) = sec A + tan A`
9 sec2 A − 9 tan2 A is equal to
Prove the following identity :
`(1 + tan^2A) + (1 + 1/tan^2A) = 1/(sin^2A - sin^4A)`
Without using trigonometric table, prove that
`cos^2 26° + cos 64° sin 26° + (tan 36°)/(cot 54°) = 2`
Prove the following identities.
tan4 θ + tan2 θ = sec4 θ – sec2 θ
Prove that `(1 + sin B)/(cos B) + (cos B)/(1 + sin B) = 2 sec B`.
If cosec θ + cot θ = p, then prove that cos θ = `(p^2 - 1)/(p^2 + 1)`
