Advertisements
Advertisements
प्रश्न
Prove the following identities.
sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1
उत्तर
L.H.S = sec4 θ (1 – sin4 θ) – 2 tan2 θ
= `1/cos^4 theta [1 - (sin^2 theta)^2]- 2 xx (sin^2 theta)/(cos^2 theta)`
= `1/(cos^4 theta) (1 + sin^2 theta) (1 - sin^2 theta) - 2 (sin^2 theta)/(cos^2 theta)`
= `1/(cos^4 theta) xx cos^2 theta (1 + sin^2 theta) - 2 (sin^2 theta)/(cos^2 theta)`
= `(1 + sin^2 theta)/(cos^2 theta) - (2sin^2 theta)/(cos^2 theta)`
= `(1 + sin^2 theta - 2sin^2 theta)/(cos^2 theta)`
= `(1 - sin^2 theta)/(cos^2 theta)`
= `(cos^2 theta)/(cos^2 theta)`
L.H.S = R.H.S
∴ sec4 θ (1 – sin4 θ) – 2 tan2 θ = 1
APPEARS IN
संबंधित प्रश्न
Prove that:
`(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)`
Prove the following identities:
cosec4 A (1 – cos4 A) – 2 cot2 A = 1
cosec4θ − cosec2θ = cot4θ + cot2θ
Prove the following identity :
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
Prove that `(tan θ + sin θ)/(tan θ - sin θ) = (sec θ + 1)/(sec θ - 1)`
Prove the following identities.
`(1 - tan^2theta)/(cot^2 theta - 1)` = tan2 θ
Prove that `(1 + sintheta)/(1 - sin theta)` = (sec θ + tan θ)2
Prove that sin θ (1 – tan θ) – cos θ (1 – cot θ) = cosec θ – sec θ
Prove that (1 – cos2A) . sec2B + tan2B(1 – sin2A) = sin2A + tan2B
The value of the expression [cosec(75° + θ) – sec(15° – θ) – tan(55° + θ) + cot(35° – θ)] is ______.