Advertisements
Advertisements
प्रश्न
Prove that `sec^2A - "cosec"^2A = (2sin^2A - 1)/(sin^2A *cos^2A)`.
Advertisements
उत्तर
L.H.S. = sec2A – cosec2A
= `1/(cos^2A) - 1/(sin^2A)`
= `(sin^2A - cos^2A)/(cos^2A*sin^2A)`
= `(sin^2A - (1 - sin^2A))/(sin^2A*cos^2A)` ...`[(∵ sin^2A + cos^2A = 1),(∴ 1 - sin^2A = cos^2A)]`
= `(sin^2A - 1 + sin^2A)/(sin^2A*cos^2A)`
= `(2sin^2"A" - 1)/(sin^2"A"*cos^2"A")`
= R.H.S.
∴ `sec^2A - "cosec"^2A = (2sin^2A - 1)/(sin^2A*cos^2A)`
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`cos theta/(1 + sin theta) = (1 - sin theta)/cos theta`
Prove the following trigonometric identities.
`(1 + cos A)/sin^2 A = 1/(1 - cos A)`
Prove the following trigonometric identities.
`(sec A - tan A)/(sec A + tan A) = (cos^2 A)/(1 + sin A)^2`
Prove the following trigonometric identities.
`1 + cot^2 theta/(1 + cosec theta) = cosec theta`
Prove the following identities:
`(sec A - 1)/(sec A + 1) = (1 - cos A)/(1 + cos A)`
Prove that:
cos A (1 + cot A) + sin A (1 + tan A) = sec A + cosec A
If` (sec theta + tan theta)= m and ( sec theta - tan theta ) = n ,` show that mn =1
If `( cos theta + sin theta) = sqrt(2) sin theta , " prove that " ( sin theta - cos theta ) = sqrt(2) cos theta`
What is the value of (1 − cos2 θ) cosec2 θ?
Prove that `sqrt((1 + sin A)/(1 - sin A))` = sec A + tan A.
Prove that `cos θ/sin(90° - θ) + sin θ/cos (90° - θ) = 2`.
Prove that: `(sin θ - 2sin^3 θ)/(2 cos^3 θ - cos θ) = tan θ`.
If cos A = `(2sqrt(m))/(m + 1)`, then prove that cosec A = `(m + 1)/(m - 1)`.
If cos A + cos2A = 1, then sin2A + sin4A = ?
The value of the expression [cosec(75° + θ) – sec(15° – θ) – tan(55° + θ) + cot(35° – θ)] is ______.
(tan θ + 2)(2 tan θ + 1) = 5 tan θ + sec2θ.
If 2sin2θ – cos2θ = 2, then find the value of θ.
If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p2 – 1) = 2p.
Let x1, x2, x3 be the solutions of `tan^-1((2x + 1)/(x + 1)) + tan^-1((2x - 1)/(x - 1))` = 2tan–1(x + 1) where x1 < x2 < x3 then 2x1 + x2 + x32 is equal to ______.
Eliminate θ if x = r cosθ and y = r sinθ.
