Advertisements
Advertisements
प्रश्न
Prove the following identities:
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2A * cos^2B)`
Advertisements
उत्तर
L.H.S. = `tan^2A - tan^2B`
= `sin^2A/cos^2A - sin^2B/cos^2B`
= `(sin^2A * cos^2B - sin^2B * cos^2A)/(cos^2A * cos^2B`
= `(sin^2A(1 - sin^2B)-sin^2B(1 - sin^2A))/(cos^2A * cos^2B)`
= `(sin^2A - sin^2A * sin^2B - sin^2B + sin^2A * sin^2B)/(cos^2A * cos^2B`
= `(sin^2A - cancel(sin^2A * sin^2B) - sin^2B + cancel(sin^2A * sin^2B))/(cos^2A * cos^2B`
= `(sin^2A - sin^2B)/(cos^2A * cos^2B)` = R.H.S.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities
(1 + cot2 A) sin2 A = 1
Prove the following trigonometric identities.
`sqrt((1 - cos theta)/(1 + cos theta)) = cosec theta - cot theta`
Prove the following trigonometric identities
`(1 + tan^2 theta)/(1 + cot^2 theta) = ((1 - tan theta)/(1 - cot theta))^2 = tan^2 theta`
Prove the following trigonometric identities.
sin2 A cos2 B − cos2 A sin2 B = sin2 A − sin2 B
If sin A + cos A = m and sec A + cosec A = n, show that : n (m2 – 1) = 2 m
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
Prove that `cos θ/sin(90° - θ) + sin θ/cos (90° - θ) = 2`.
If `(cos alpha)/(cos beta)` = m and `(cos alpha)/(sin beta)` = n, then prove that (m2 + n2) cos2 β = n2
Prove that sin6A + cos6A = 1 – 3sin2A . cos2A.
If 1 + sin2θ = 3 sin θ cos θ, then prove that tan θ = 1 or `1/2`.
