Advertisements
Advertisements
Question
Prove the following identities.
tan4 θ + tan2 θ = sec4 θ – sec2 θ
Advertisements
Solution
tan4 θ + tan2 θ = sec4 θ – sec2 θ
L.H.S = tan4 θ + tan2 θ
Taking out tan2 θ as common
= tan2 θ (tan2 θ + 1)
We know that
1 + tan2 θ = sec2 θ
i.e. tan2 θ = sec2 θ - 1
It can be written as
= (sec2 θ – 1) sec2 θ
So we get
= sec4 θ – sec2 θ
= R.H.S
Therefore, it is proved.
APPEARS IN
RELATED QUESTIONS
If (secA + tanA)(secB + tanB)(secC + tanC) = (secA – tanA)(secB – tanB)(secC – tanC) prove that each of the side is equal to ±1. We have,
Prove the following trigonometric identities.
`(1 + cos θ + sin θ)/(1 + cos θ - sin θ) = (1 + sin θ)/cos θ`
Prove the following identities:
`(1 - sinA)/(1 + sinA) = (secA - tanA)^2`
Prove that:
(tan A + cot A) (cosec A – sin A) (sec A – cos A) = 1
`(sec^2 theta-1) cot ^2 theta=1`
Prove the following identity :
`(cotA + cosecA - 1)/(cotA - cosecA + 1) = (cosA + 1)/sinA`
Prove the following identity :
`(cosecθ)/(tanθ + cotθ) = cosθ`
If `1 - cos^2θ = 1/4`, then θ = ?
Prove that `(cos^2θ)/(sinθ) + sin θ = "cosec" θ`.
The value of tan A + sin A = M and tan A - sin A = N.
The value of `("M"^2 - "N"^2) /("MN")^0.5`
