Advertisements
Advertisements
Question
Prove the following identity :
`(cos^3θ + sin^3θ)/(cosθ + sinθ) + (cos^3θ - sin^3θ)/(cosθ - sinθ) = 2`
Advertisements
Solution
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
RELATED QUESTIONS
Prove the following trigonometric identities.
sec6 θ = tan6 θ + 3 tan2 θ sec2 θ + 1
Prove the following trigonometric identities.
`(sec A - tan A)/(sec A + tan A) = (cos^2 A)/(1 + sin A)^2`
If cos θ + cot θ = m and cosec θ – cot θ = n, prove that mn = 1
Prove the following identity :
`sqrt((1 + cosA)/(1 - cosA)) = cosecA + cotA`
Prove the following identity :
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2Acos^2B)`
If m = a secA + b tanA and n = a tanA + b secA , prove that m2 - n2 = a2 - b2
Find the value of x , if `cosx = cos60^circ cos30^circ - sin60^circ sin30^circ`
Prove that:
`sqrt((sectheta - 1)/(sec theta + 1)) + sqrt((sectheta + 1)/(sectheta - 1)) = 2cosectheta`
Prove the following identities:
`1/(sin θ + cos θ) + 1/(sin θ - cos θ) = (2sin θ)/(1 - 2 cos^2 θ)`.
If cos A + cos2A = 1, then sin2A + sin4 A = ?
