Advertisements
Advertisements
प्रश्न
Prove that:
`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`
Advertisements
उत्तर
LHS = `(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos^3 θ - sin^3 θ)/(cos θ - sin θ)`
= `((cos θ + sin θ)(cos^2 θ + sin^2 θ - cos θ sin θ))/(cos θ + sin θ) + ((cos θ - sin θ)(cos^2 θ + sin^2 θ - cos θ sin θ))/(cos θ - sin θ)`
= 1 - sin θ cos θ + 1 + sin θ cos θ
= 2
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities:
(i) (1 – sin2θ) sec2θ = 1
(ii) cos2θ (1 + tan2θ) = 1
Prove the following trigonometric identity:
`sqrt((1 + sin A)/(1 - sin A)) = sec A + tan A`
`(sin theta)/((sec theta + tan theta -1)) + cos theta/((cosec theta + cot theta -1))=1`
`(tan A + tanB )/(cot A + cot B) = tan A tan B`
If `cos theta = 7/25 , "write the value of" ( tan theta + cot theta).`
Prove the following identity :
`(1 - cos^2θ)sec^2θ = tan^2θ`
Prove the following identity :
`(cosecA)/(cosecA - 1) + (cosecA)/(cosecA + 1) = 2sec^2A`
If tan α = n tan β, sin α = m sin β, prove that cos2 α = `(m^2 - 1)/(n^2 - 1)`.
Eliminate θ if x = r cosθ and y = r sinθ.
Proved that `(1 + secA)/secA = (sin^2A)/(1 - cos A)`.
