Advertisements
Advertisements
प्रश्न
Prove the following identities.
tan4 θ + tan2 θ = sec4 θ – sec2 θ
Advertisements
उत्तर
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
संबंधित प्रश्न
Prove the following trigonometric identities.
(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1
`(sin theta)/((sec theta + tan theta -1)) + cos theta/((cosec theta + cot theta -1))=1`
If `sec theta = x ,"write the value of tan" theta`.
If sin2 θ cos2 θ (1 + tan2 θ) (1 + cot2 θ) = λ, then find the value of λ.
Prove that:
(cosec θ - sinθ )(secθ - cosθ ) ( tanθ +cot θ) =1
Prove the following identity :
tanA+cotA=secAcosecA
Prove the following identity :
`(1 + cosA)/(1 - cosA) = tan^2A/(secA - 1)^2`
Prove that (sin θ + cosec θ)2 + (cos θ + sec θ)2 = 7 + tan2 θ + cot2 θ.
Prove that `((1 + sin θ - cos θ)/( 1 + sin θ + cos θ))^2 = (1 - cos θ)/(1 + cos θ)`.
Prove the following identities: cot θ - tan θ = `(2 cos^2 θ - 1)/(sin θ cos θ)`.
