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 that `(sin theta)/(1-cottheta) + (cos theta)/(1 - tan theta) = cos theta + sin theta`
Prove the following trigonometric identities.
`[tan θ + 1/cos θ]^2 + [tan θ - 1/cos θ]^2 = 2((1 + sin^2 θ)/(1 - sin^2 θ))`
Prove the following trigonometric identities
sec4 A(1 − sin4 A) − 2 tan2 A = 1
Prove the following trigonometric identities.
`(1 + cot A + tan A)(sin A - cos A) = sec A/(cosec^2 A) - (cosec A)/sec^2 A = sin A tan A - cos A cot A`
If sec A + tan A = p, show that:
`sin A = (p^2 - 1)/(p^2 + 1)`
`(tan^2theta)/((1+ tan^2 theta))+ cot^2 theta/((1+ cot^2 theta))=1`
`sqrt((1 + sin θ)/(1 - sin θ)) = sec θ + tan θ`
Write the value of tan1° tan 2° ........ tan 89° .
What is the value of (1 − cos2 θ) cosec2 θ?
If x = a sin θ and y = b cos θ, what is the value of b2x2 + a2y2?
If \[\sin \theta = \frac{4}{5}\] what is the value of cotθ + cosecθ?
If sec θ + tan θ = x, then sec θ =
Prove the following identity :
cosecθ(1 + cosθ)(cosecθ - cotθ) = 1
Prove the following Identities :
`(cosecA)/(cotA+tanA)=cosA`
Prove the following identity :
`(1 + tan^2θ)sinθcosθ = tanθ`
If sinA + cosA = m and secA + cosecA = n , prove that n(m2 - 1) = 2m
Proved that cosec2(90° - θ) - tan2 θ = cos2(90° - θ) + cos2 θ.
If A + B = 90°, show that sec2 A + sec2 B = sec2 A. sec2 B.
If `sqrt(3)` sin θ – cos θ = θ, then show that tan 3θ = `(3tan theta - tan^3 theta)/(1 - 3 tan^2 theta)`
If sin θ + cos θ = p and sec θ + cosec θ = q, then prove that q(p2 – 1) = 2p.
