Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(cos^3θ + sin^3θ)/(cosθ + sinθ) + (cos^3θ - sin^3θ)/(cosθ - sinθ) = 2`
Advertisements
उत्तर
LHS = `(cos^3θ + sin^3θ)/(cosθ + sinθ) + (cos^3θ - sin^3θ)/(cosθ - sinθ)`
= `((cos^3θ + sin^3θ)(cosθ - sinθ) + (cos^3θ - sin^3θ)(cosθ + sinθ))/((cosθ + sinθ)(cosθ - sinθ))`
= `(cos^4θ - cos^3θsinθ + sin^3θcosθ - sin^4θ + cos^4θ + cos^3θsinθ - sin^3θcosθ - sin^4θ)/(cos^2θ - sin^2θ)`
= `(2cos^4θ - 2sin^4θ)/(cos^2θ - sin^2θ) = (2(cos^4θ - sin^4θ))/(cos^2θ - sin^2θ)`
= `(2(cos^2θ + sin^2θ)(cos^2θ - sin^2θ))/((cos^2θ - sin^2θ))` = 2(`cos^2θ + sin^2θ`)
= 2 `(∵(cos^2θ + sin^2θ) = 1)`
OR
LHS = `(cos^3θ + sin^3θ)/(cosθ + sinθ) + (cos^3θ - sin^3θ)/(cosθ - sinθ)`
= `((cosθ + sinθ)(cos^2θ + sin^2θ - cosθ sinθ))/(cosθ + sinθ) + ((cosθ - sinθ)(cos^2θ + sin^2θ + cosθsinθ))/((cosθ - sinθ))` (∵ `a^3 ± b^3 = (a ± b)(a^2 + b^2 ± ab`))
= `(cos^2θ + sin^2θ - cosθsinθ) + (cos^2θ + sin^2θ + cosθsinθ)`
= `1 - cosθsinθ + 1 + cosθsinθ` (∵ `cos^2θ + sin^2θ = 1`)
= 2
APPEARS IN
संबंधित प्रश्न
`Prove the following trigonometric identities.
`(sec A - tan A)^2 = (1 - sin A)/(1 + sin A)`
If 3 sin θ + 5 cos θ = 5, prove that 5 sin θ – 3 cos θ = ± 3.
Prove the following identities:
(cosec A – sin A) (sec A – cos A) (tan A + cot A) = 1
`(1+tan^2theta)(1+cot^2 theta)=1/((sin^2 theta- sin^4theta))`
cos4 A − sin4 A is equal to ______.
Prove the following identity :
`tanA - cotA = (1 - 2cos^2A)/(sinAcosA)`
If x = r sinA cosB , y = r sinA sinB and z = r cosA , prove that `x^2 + y^2 + z^2 = r^2`
If tanA + sinA = m and tanA - sinA = n , prove that (`m^2 - n^2)^2` = 16mn
If cot θ + tan θ = x and sec θ – cos θ = y, then prove that `(x^2y)^(2/3) – (xy^2)^(2/3)` = 1
Prove that `sintheta/(sectheta+ 1) +sintheta/(sectheta - 1)` = 2 cot θ
