Advertisements
Advertisements
प्रश्न
Prove the following identity :
`sin^2Acos^2B - cos^2Asin^2B = sin^2A - sin^2B`
Advertisements
उत्तर
LHS = `sin^2A(1 - sin^2B) - (1 - sin^2A)sin^2B`
= `sin^2A - sin^2A.sin^2B - sin^2B + sin^2A.sin^2B`
= `sin^2A - sin^2B` = RHS
APPEARS IN
संबंधित प्रश्न
(1 + tan θ + sec θ) (1 + cot θ − cosec θ) = ______.
`(1+tan^2A)/(1+cot^2A)` = ______.
Prove the following trigonometric identities.
sin2 A cot2 A + cos2 A tan2 A = 1
Prove the following identities:
`(cotA - cosecA)^2 = (1 - cosA)/(1 + cosA)`
Prove that `( sintheta - 2 sin ^3 theta ) = ( 2 cos ^3 theta - cos theta) tan theta`
If `x/(a cosθ) = y/(b sinθ) "and" (ax)/cosθ - (by)/sinθ = a^2 - b^2 , "prove that" x^2/a^2 + y^2/b^2 = 1`
Prove that `(sin (90° - θ))/cos θ + (tan (90° - θ))/cot θ + (cosec (90° - θ))/sec θ = 3`.
Prove that cos θ sin (90° - θ) + sin θ cos (90° - θ) = 1.
Prove that `(cos(90 - "A"))/(sin "A") = (sin(90 - "A"))/(cos "A")`
If sinθ – cosθ = 0, then the value of (sin4θ + cos4θ) is ______.
