Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA) = 2`
उत्तर
LHS = `(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA)`
= `((cos^3A + sin^3A)(cosA - sinA) + (cos^3A - sin^3A)(cosA + sinA))/(cos^2A - sin^2A)`
= `(cos^4A - cos^3AsinA + sin^3AcosA - sin^4A + cos^4A + cos^3AsinA - sin^3AcosA = sin^4A)/(cos^2A - sin^2A)`
= `(2(cos^4A - sin^4A))/(cos^2A - sin^2A) = (2(cos^2A + sin^2A)(cos^2A - sin^2A))/((cos^2A - sin^2A)) = 2(cos^2A + sin^2A)`
= `2(∵ cos^2A + sin^2A = 1)`
OR
= `(cos^3A + sin^3A)/(cosA + sinA) + (cos^3A - sin^3A)/(cosA - sinA)`
= `((cosA + sinA)(cos^2A + sin^2A - cosAsinA))/((cosA + sinA)) + ((cosA - sinA)(cos^2A + sin^2A + cosAsinA))/((cosA - sinA))` (∵ a3 ± b3 = (a ± b)(a2 + b2 ± ab))
= `(cos^2A + sin^2A - cosAsinA) + (cos^2A + sin^2A + cosAsinA)`
= `1 - cosAsinA + 1 + cosAsinA ` (∵ `cos^2A + sin^2A = 1`)
= 2
APPEARS IN
संबंधित प्रश्न
The angles of depression of two ships A and B as observed from the top of a light house 60 m high are 60° and 45° respectively. If the two ships are on the opposite sides of the light house, find the distance between the two ships. Give your answer correct to the nearest whole number.
Prove the following trigonometric identities.
`(sec A - tan A)/(sec A + tan A) = (cos^2 A)/(1 + sin A)^2`
`(sec^2 theta -1)(cosec^2 theta - 1)=1`
If `( tan theta + sin theta ) = m and ( tan theta - sin theta ) = n " prove that "(m^2-n^2)^2 = 16 mn .`
If cos A + cos2 A = 1, then sin2 A + sin4 A =
Prove the following identity :
`(tanθ + secθ - 1)/(tanθ - secθ + 1) = (1 + sinθ)/(cosθ)`
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`
`(sin A)/(1 + cos A) + (1 + cos A)/(sin A)` = 2 cosec A
If a cos θ – b sin θ = c, then prove that (a sin θ + b cos θ) = `± sqrt("a"^2 + "b"^2 -"c"^2)`
Choose the correct alternative:
1 + cot2θ = ?
